C4

 C4 Chemical Changes Mastery Booklet ANSWERSPart 1: Formulae of ionic compounds (Recap questions)Beryllium atomOxygen atomThe two outer electrons from Beryllium are transferred to the oxygen atom when Beryllium reacts with oxygen.The two ions formed in this reaction are...Beryllium ionOxide IonWhen Lithium reacts with oxygen, the outer electrons from two Lithium atoms are transferred to one oxygen atom.Lithium AtomOxygen atomLithium ionOxide IonLithium oxide has the formula Li2O but the formula for beryllium oxide is BeO because a Lithium atom has one electron in its outer sheet. Oxygen is able to receive two electrons to achieve Noble Gas configuration so two Lithium atoms are needed. Beryllium is in group 2 of the periodic table. Its outer shell contains two electrons so only one Beryllium atom is needed to give oxygen the two electrons it needs.Challenge: Predict the formulae of:Lithium fluoride LiFLithium nitride Li3NAluminium nitride AlNAluminium chloride AlCl3Aluminium oxide Al2O3Magnesium nitride Mg3N2PracticeCalcium chloride CaCl2Calcium sulfate CaSO4Lithium sulfate Li2SO4Lithium nitrate LiNO3Aluminium bromide AlBr3Aluminium oxide Al2O3Aluminium nitrate Al(NO3)3Aluminium sulfate Al2(SO4)3Ammonium iodide NH4IAmmonium sulfate (NH4)2SO4Iron (II) sulfate FeSO4Iron (III) sulfate Fe2(SO4)3Sodium nitrate NaNO3Barium sulfate (think about what group it is in!) BaSO4Copper (I) hydroxide CuOHCopper (I) sulfate Cu2SO4Ammonium oxide (NH4)2OAmmonium carbonate (NH4)2CO3Aluminium hydroxide Al(OH)3Copper (II) carbonate CuCO3Beryllium fluoride BeF2Beryllium sulfate BeSO4Vanadium (V) carbonate V2(CO3)5Manganese (VII) iodide MnI7Manganese (VII) sulphate Mn2 (SO4)7Part 2: Reactivity of Metals 5185410163830Recap questions:Metallic bonding can be described as the sharing of delocalised (free) electrons among a lattice of positively charged metal ions.(1) Magnesium will conduct electricity, (2) malleable, (3) ductile.Magnesium conducts electricity as the delocalised electrons move through the structure carrying charge.Word equation Francium + Bromine → Francium BromideSymbol equation2Fr + Br2 → 2FBr100g of Fr is used with 50g of bromine. Which is excess and which is limiting?Fr moles = mass /Mr = 100/223 = 0.00448 molesBr2 moles = mass / Mr = 50/(79.9 x 2)= 0.313 molesFr:Br2 react in a 2:1 ratio. Therefore 0.00448 moles of Francium will react with 0.00224 moles of Bromine. Bromine is in excess. Francium is the limiting reagent.75g of Fr is reacted with an excess of bromine. How much FrBr is formed?nFr = m/Mr = 75/223 = 0.336 molesFr:Br2 react in a 2:1 ratio. Therefore 0.336 moles of Francium will react with 0.168 moles of Bromine. mBr2 = n x Mr = 0.168 x 159.8 = 26.9g (3.s.f.)When FrBr is molten or dissolved in water the ions are free to move through the substance, carrying charge.Fr is more reactive than Li because a Fr atom is larger, meaning its electrons are further from the nucleus. The force of attraction between the electrons and the positive protons in the nucleus is weaker so the electrons are more easily lost making Francium more reactive.There are two isotopes of Francium, Fr-223 and Fr-221. Making reference to their numbers of protons, neutrons and electrons, discuss the similarities and differences between these two isotopes.Similarities- Atomic number 87. 87 protons. 87 electrons.Differences- Mass numbers. Fr-223 has 136 neutrons whereas Fr-221 has 134 neutrons.Average mass of the atoms= ((81 x 223)+ (19 x 221))/100 = 180.63 + 41.99 = 222.62 = 223 (3 s.f.)Challenge: (70 x 223) + (22 x 221) + (8 x X) /100 = 222.32156.1 + 48.62 + 0.08X = 222.320.08X = 222.32- 156.1 - 48.62 0.08X = 17.6 X = 17.6/ 0.08 = 220 Magnesium is more reactive than copper as Magnesium atoms are smaller than copper atoms so the outer electrons of Magnesium are more easily lost.Magnesium + oxygen → Magnesium Oxide2 Mg (s) + O2 (g) → 2 MgO (s)Reactants → ProductsA reaction where oxygen is added is called an oxidation reaction. If oxygen is removed, it is called a reduction reaction.Magnesium atomOxygen atomUse arrows to show what occurs when they react with each other (Arrow from two outer electrons of Magnesium to outer shell of Oxygen)Draw the ions that are formed as a result of this reaction.Magnesium ionOxide IonExplain why this reaction is called an oxidation reaction. Oxygen is added to Magnesium. Oxidation involves a loss of electrons. The Magnesium atom is oxidised to form the Magnesium ion. The mass increases because Oxygen from the air was added.Magnesium oxide → Magnesium + OxygenDecompositionOxygen has a low melting point as it exists as small simple molecules. There are weak forces of attraction between the molecules that are easily overcome with little heat energy at low temperatures.Magnesium has a high melting point as the metallic bonds are strong and require lots of energy to break.Magnesium oxide will conduct electricity when it is molten or dissolved in water (aqueous)Challenge: Calcium is more reactive than Magnesium as calcium atoms are larger therefore the outer electrons are further from the nucleus and the attraction between them and the protons is weaker. This means that the outer electrons are more easily lost.Magnesium> Iron> CopperWhen Iron atoms react they lose 3 electrons to form Fe3+ ions.The type, concentration and volume of acid are control variables. Strong acids react differently to weak acids. You would also see more of a reaction with a more concentrated acid than a more dilute acid.The student should also use the same mass of metal, keep the same surface area (i.e. all powder or all 2cm x 2cm pieces) and keep the acid at the same temperature.In Fe2(SO4)3 there are 17 atoms; 2 iron, 3 sulfur and 12 oxygen.Challenge: smaller metals with fewer electrons react more vigorously than metals with larger atoms.More reactive> less reactiveCalcium < lithiumGold < copperSodium >ironZinc > copperCopper <zincIron < zincIron < calciumSodium > lithiumFor each reaction below, state whether or not it would occur.Magnesium oxide + calcium NOIron chloride + zinc YESCopper bromide + gold NOZinc chloride + potassium YESIron sulphate + copperNOIron + lithium sulphateNOMagnesium + iron oxideYESPotassium + Sodium Chloride → Sodium + Potassium ChloridePotassium is more reactive than sodium so displaced the sodium from the compound.Zinc + Iron Bromide → Zinc Bromide + IronElements- Zinc, iron, bromine. Compounds- Iron Bromide and Zinc BromideOnly more reactive metals displace less reactive metals.Copper is less reactive than Lithium so cannot displace it from the compound.CaCl2 + 2 Li ? 2 LiCl + CaCalcium ions have a +2 . Formula= Ca2+ElementProtonElectronNeutronCalcium202020Chlorine-35171718Lithium334Subatomic particleProtonElectronNeutronCharge+1-10Mass11/1000 or very small (negligible)1Always using 1 minute is a control variable. It ensures the results can be compared. i.e. fair test.Y> Z > XMetal Y would react easiest with oxygen.Alloys are mixtures not compounds so Metal X will still have three electrons in its outer shell as it will not form an ion.Part 3: Extracting metalsOxygen is removed from iron oxide leaving the iron as an element. Reduction is removal of oxygen.An Ore has to have enough metal in it to be economically viable to extract (ie. make a profit). It mustn’t contain enough of the metal to make a profit.Francium can only be extracted using electrolysis.How reactive it was compared to Carbon and (Hydrogen).Lithium would react the most because it is a group 1 metal. Group 1 metals are more reactive than group 2, 3 and transition metals. Zinc Oxide + carbon → Zinc + Carbon DioxideGold is found naturally in the Earth’s crust as it is quite unreactive.Silver would go above gold but near the bottom of the reactivity series.Lithium + copper oxide→ lithium oxide + copper2Li + CuO ? Li2O + CuLithium has been oxidised as oxygen has been added to form a compound and copper oxide has been reduced as oxygen has been removed to leave copper as an element. Iron oxide + sodium → sodium oxide + iron. Fe2O3 + 6 Na ? 3 Na2O + 2 Fethe iron in iron oxide has been reduced to form iron. Sodium has been oxidised to form sodium oxide.Zinc and Potassium OxidePotassium has been oxidised to Potassium Oxide and the Zinc in Zinc Oxide has been reduced to Zinc.. GCSE Practice Questions?MetalTemperature rise in °CMeantemperaturerise in °CTest 1Test 2Test 3Calcium17.816.917.5?Iron??6.2??6.0??6.1??6.1Magnesium12.5??4.212.312.4Zinc??7.8??8.0??7.6??7.8(a)?????Volume and concentration of acid. Size/ surface area and mass of metal used.(b)?????4.2 is anomalous as it is much smaller than the other results. The temperature may have been measured wrongly or the mass of magnesium may have been smaller or a smaller volume of acid may have been used.(c)?????Mean temp rise = (17.8 + 16.9 + 17.5 ) /3 = 17.4(e)?????The temperature rise when aluminium is reacted with dilute hydrochloric acid would be about 10 degrees.Question 2: A student investigated the reactivity of three different metals. This is the method used.1.???????Place 1 g of metal powder in a test tube.2.???????Add 10 cm3 of metal sulfate.3.???????Wait 1 minute and observe.4.???????Repeat using the other metals and metal sulfates.The student placed a tick in the table below if there was a reaction and a cross if there was no reaction.??ZincCopperMagnesiumCopper sulfateMagnesium sulfateZinc sulfate(a)?????The dependent variable in the investigation is the reaction.(b) - no question(c)?????Mass of metal → Balance, Volume of metal sulfate → Measuring cylinder/ Burette? (d)????Magnesium> Zinc> Copper (e)?????Sodium is very reactive so this could be unsafe (f)????? Gold is found in the Earth as the metal itself. (g)????2 Fe2O3????? +?????3 C????? →?????4 Fe????? +????? 3 CO2(h)?? The element Carbon is used to reduce iron oxide.(i)????Reduction is removal of oxygen (or a gain of electrons)Question 3: Metals are used in the manufacture of pylons and overhead power cables.?(a) Iron (steel) is used to make pylons because it conducts electricity.(b) ????The table shows some of the properties of two metals. Use the information in the table to suggest why aluminium and not copper is used to conduct electricity in overhead power cables.?MetalDensity in g per cm3Melting point in°CPercentage(%) relative electrical conductivityPercentage(%) abundance in Earth’s crustcopper8.9210831000.007aluminium2.70660608.1Copper is heavy as its density is high and it has a low abundance so there isn’t much available to use. Aluminium also has a low percentage abundance. It’s also very soft with a low melting point. It only has a 60% conductivity so won’t conduct electricity as well.Part 4: Acids1. Reactions of acids with metalsFormula of saltName of saltOriginal metalOriginal acidNaClSodium ChlorideSodiumHydrochloric AcidLiClLithium ChlorideLithiumHydrochloric AcidCaSO4Calcium SulfateCalciumSulfuric AcidMgSO4Magnesium SulfateMagnesiumSulfuric AcidKNO3Potassium NitratePotassiumNitric AcidWrite a word equation for the reactions between:Aluminium + Hydrochloric acid → Aluminium Chloride + HydrogenMagnesium + Hydrochloric acid → Magnesium Chloride + HydrogenCalcium + Nitric acid → Calcium Nitrate + HydrogenBeryllium + Sulfuric acid → Beryllium Sulfate + HydrogenIron + Sulfuric acid → Iron Sulfate + HydrogenWrite a word equation for a reaction which forms:Aluminium + Sulfuric Acid → Aluminium Sulfate + HydrogenCalcium + Hydrochloric Acid → Calcium Chloride + HydrogenBarium + Nitric Acid → Barium Nitrate + HydrogenRubidium + Hydrochloric Acid → Rubidium Chloride + HydrogenChallenge: write symbol equations for every reaction in Q1 and Q2Write a symbol equation:2 Al + 6 HCl → 2 AlCl3 + 3 H2Mg + 2 HCl → MgCl2 + H2Ca + HNO3 → Ca(NO3)2 + H2Be + H2SO4 → BeSO4 + H22 Fe + H2SO4 → Fe2SO4 + H2Write a symbol equation:Al + H2SO4 → Al2 (SO4)3 + H2Cal + 2 HCl → CaCl2 + H2Ba + HNO3 → Ba(NO3)2 + H22 Rb+ 2 HCl →2 RbCl + H2Part 5: Ionic equationsComplete the table below.AtomIonAtomIonAtomIonLiLi+BeBe2+AlAl3+NaNa+MgMg2+Fe (II)Fe2+KK+CaCa2+Fe (III)Fe3+For each of the salts formed in question 2 and question 3 write out the symbol formula- Already done plete the equations:Mg + H2SO4 ? MgSO4 + H22Li + 2HNO3 ? 2 LiNO3 + H22Al + 6HNO3?2 Al(NO3)3+ 3H22Fe(III) + 6HCl ? 2 FeCl3 + 3H2For the equations below, you will need to balance and complete:Mg + HNO3 ? Mg (NO3)2+ H2Fe(III) + HNO3 ? Fe (NO3)3+ H2K + H2SO4 ? K2SO4 + H2Li + H2SO4 ? Li2SO4+ H2Write out full word and symbol equations for each of the below:Iron (II) + Nitric acid → Iron Nitrate + HydrogenFe + 2 HNO3 → Fe(NO3)2 + H2Barium + Sulfuric acid → Barium Sulfate + HydrogenBa + H2SO4 → BaSO4 + H2Strontium + Hydrochloric acid → Strontium Chloride + HydrogenSr + 2 HCl → SrCl2 + H2Aluminium + Sulfuric acid → Aluminium Sulfate2 Al + 3 H2SO4 → Al2(SO4)3 + 3 H2Gallium + Nitric acid → Gallium Nitrate + Hydrogen2 Ga + 6HNO3 → 2 Ga(NO3)3 + 3 H2Part 5: Ionic equationsComplete the ionic equations:s. Mg (s) + 2H+ (aq) ? Mg2+ (aq) + H2 (g)t. Li (s) + 2H+ (aq) ? 2 Li+ (aq) + H2u. 2Al (s) + 6H+ (aq) ?2 Al3+ (aq) + 3H2 (g)v. 2Fe(s) + 6H+ (aq) ? 2 Fe3+ (aq) + 3H2 (g)Ionic equations:w. Mg (s) + H+ (aq) ? Mg2+(aq) + H2 (g)x. Fe(s) + H+ (aq) ? Fe 3+ (aq) + H2 (g)y. K(s) + 2 H+ (aq) ? 2 K+ (aq) + H2 (g)z. Li(s) + 2 H+ (aq) ? 2 Li + (aq)+ H2 (g)Ionic equations;aa. Fe (s) + 2 H+ (aq) → Fe2+ (aq) + H2 (g)bb. Ba (s) + 2 H+ (aq) → Ba2+ (aq) + H2 (g)cc. Sr(s) + 2 H+ (aq) → Sr2+ (aq)+ H2 (g)dd. 2 Al(s) + 6 H+ (aq) → 2 Al3+ (aq) + 3 H2 (g)ee. 2 Ga(s) + 6H+ (aq)→ 2 Ga3+ (aq) + 3 H2 (g)Part 6: Half equationsFor each of the ionic equations in question 106 construct half equations. aa. Fe (s)) → Fe2+ (aq) + 2 e- bb. Ba (s) → Ba2+ (aq) + 2 e- cc. Sr(s) → Sr2+ (aq)+ 2 e- dd. 2 Al(s) → 2 Al3+ (aq) + 6 e- ee. 2 Ga(s) → 2 Ga3+ (aq) + 6 e- Acids continued: 5. Metal HydroxidesState the formulae for the hydroxides of: lithium, potassium, calcium, aluminium, magnesium, iron (II), iron (III), zinc (II) and beryllium→ LiOH, KOH, Ca(OH)2, Al(OH)3, Mg(OH)2, Fe(OH)2, Fe(OH)3, Zn(OH)2, Be(OH)26. Alkali + acid ? salt + waterFor each of the reactions below, write:A word equationA balanced symbol equation (assume that all hydroxides are dissolved in water)An ionic equationPotassium Hydroxide + Hydrochloric Acid→ Potassium Chloride + WaterKOH + HCl → KCl + H2OH+(aq) + OH- (aq)→ H2O (l)Magnesium Hydroxide + Hydrochloric Acid→ Magnesium Chloride + WaterMg(OH)2 + 2 HCl → MgCl2 + 2 H2OH+(aq) + OH- (aq)→ H2O (l)Sodium Hydroxide + Nitric Acid → Sodium Nitrate + WaterNaOH + HCl → NaCl + H2OH+(aq) + OH- (aq)→ H2O (l)Sodium hydroxide and sulphuric acidNaOH + H2SO4 → Na2SO4 + H2OH+(aq) + OH- (aq)→ H2O (l)Iron (III) hydroxide and sulphuric acidFe(OH)3 + H2SO4 → Fe2(SO4)3 + H2OH+(aq) + OH- (aq)→ H2O (l)7. Acid and baseWrite the formula for the oxides of lithium, potassium, calcium, aluminium, magnesium, iron (II), iron (III), zinc (II) and beryllium→ Li2O, K2O, CaO, Al2O3, MgO, FeO, Fe2O3, ZnO, BeO8. Metal oxide + acid ? salt + waterPotassium Oxide + Hydrochloric Acid → Potassium Chloride + WaterK2O + 2 HCl → 2 KCl + H2OMagnesium Oxide + Hydrochloric Acid → Magnesium Chloride + WaterMgO + 2 HCl → MgCl2 + H2OSodium oxide + Nitric Acid → Sodium Nitrate + WaterNa2O + 2 HNO3 → 2 NaNO3 + H2OSodium oxide + Sulfuric acid → Sodium Sulfate + WaterNa2O + H2SO4 → Na2SO4 + H2OIron (III) oxide + sulphuric acid → Ion (III) Sulfate + WaterFe2O3 + 3 H2SO4 → Fe2(SO4)3 + 3 H2O9. Metal carbonate + acid ? salt + water + carbon dioxideState the formulae for the carbonates of: lithium, potassium, calcium, aluminium, magnesium, iron (II), iron (III), zinc (II) and beryllium→ Li2CO3, K2CO3, CaCO3, Al2(CO3)3, MgCO3, FeCO3, Fe2(CO3)3, ZnCO3, BeCO3Potassium Carbonate + Hydrochloric Acid → Potassium Chloride + Water + Carbon DioxideK2CO3 + 2 HCl → 2 KCl + H2O + CO2 Magnesium Carbonate + Hydrochloric Acid → Magnesium Chloride + Water + Carbon DioxideMgCO3 + 2 HCl → MgCl2 + H2O + CO2 Sodium Carbonate + Nitric Acid → Sodium Nitrate + Water + Carbon DioxideNa2CO3 + 2 HNO3 → 2 NaNO3 + H2O + CO2 Sodium Carbonate + Sulfuric Acid → Sodium Sulfate + Water + Carbon DioxideNa2CO3 + H2SO4 → Na2SO4 + H2O + CO2 Iron (III) Carbonate + Sulfuric Acid → Iron (III) Sulfate + Water + Carbon DioxideFe2(CO3) 3 + 3 H2SO4 → Fe2(SO4)3 + 3 H2O + 3 CO2 pHQuestions:A substance has 40g of acid in it and is pH 2.What mass of acid would be required for pH 1? 400gWhat mass of acid would be required for pH 3? 4gWhat mass of acid would be required for pH 4? 0.4gA certain solution has 12050 H+ ions in it and has a pH of 3.How many H+ ions would need to be added for a pH of 1? x10x10 = x100 added to move 2 x pH units. so 1, 205, 000 would be in solution. 1, 205, 000- 12, 050 = 1,192,950 additional H+ ions would need to be added.How many H+ ions would need to be taken away for a pH of 5? pH would contain 120.5 H+ ions, say 121 as you can’t have 0.5 of an H+ ion. so 12050-120.5 = 11, 929 H+ ions would need to be removed.Strong, weak, dilute, concentratedSulphuric acid is a strong acid as it fully ionises in water.Sulphuric acid split up into 2 x hydrogen ions (H+) and one SO42- ion. 20g sulfuric acid would be needed to make a pH of 1Complete the table below, using the words “high,” “medium” or “low” to represent pH. The first one has been done for you.ConcentratedDiluteStrong acidLow MediumWeak acidMediumHighSulphuric acid + sodium hydroxide → Sodium Sulfate + WaterH2SO4 (aq) + 2 NaOH(aq) → Na2SO4 (aq) + 2 H2O (aq)H+ (aq) + OH- (aq) → H2O (l)If 10g of acid is dissolved in water, followed by another 10g the pH will reduce by 2.A strong acid has a lower pH than a weak acid as there are more H+ ions in solution for strong acids as they fully dissociate. Weak acids only partially dissociate so there are fewer H+ ions in a solution of weak acid.Back to metal extractionIn a reaction, solid sodium displaces iron from iron (II) sulphate solution:Na + FeSO4 ? Na2SO4 + FeBalance the equation 2 Na + FeSO4 ? Na2SO4 + FeAdd state symbols to the equation 2 Na (s) + FeSO4 (aq) ? Na2SO4 (aq) + Fe (s)Write an ionic equation for this reaction 2 Na + Fe2+ → 2 Na+ + FeUse half equations to show that sodium has been oxidised and iron ions have been reducedNa → Na+ + e- A Sodium atom has been oxidised as it has lost an electron to form a sodium ion.Fe2+ + 2 e- → Fe Iron ions have been reduced to Fe atomsRepeat the above process for reactions between:Na(s) + LiOH(aq) ? NaOH(aq) + Li(s)Na(s) + LiOH(aq) ? NaOH(aq) + Li(s) BalancedIonic equation: Na + Li+ → Na+ + LiHalf equations: Na → Na+ + e- and Li+ + e- → LiCa(s) + MgCl2(aq) ? CaCl2(aq) + Mg(s)Ca(s) + MgCl2(aq) ? CaCl2(aq) + Mg(s) BalancedIonic equation: Ca + Mg2+ ? Ca2+ + MgHalf equations: Ca → Ca2+ + 2 e- and Mg2+ + 2 e- → MgCa(s) + FeSO4(aq) ? CaSO4(aq) + Fe(s) (the charge on the iron ion in iron sulphate is 2+)Ca(s) + FeSO4(aq) ? CaSO4(aq) + 2Fe(s) BalancedIonic equation: Ca + Fe2+? Ca2+ + Fe(s) Half equations: Ca → Ca2+ + 2 e- and Fe2++ 2 e- ? Fe(s) Sodium is more reactive than lithium because a sodium atom is larger than a lithium atom so it’s outer electrons are not held as strongly. The attraction between them and the protons in the nucleus is weaker so they are more easily lost, making sodium more reactive.Repeat the process above for the reaction:LiCl(aq) + F2(aq) ? LiF(aq) + Cl2 (g)2 LiCl(aq) + F2(aq) ? 2 LiF(aq) + Cl2 (g)Ionic equation: 2 Cl- + F2 → Cl2 + 2F-Half equations: 2 Cl- → Cl2 + 2 e- and F2 + 2 e- → 2F-Repeat the process above for the reaction:MgI2(aq) + Br2(aq) ? MgBr2(aq) + I2(aq)MgI2(aq) + Br2(aq) ? MgBr2(aq) + I2(aq) BalancedIonic equation: 2I- → I2 + 2e- and Br2 + 2e- → 2Br-There is no reaction between sodium chloride and bromine because bromine is less reactive than chlorine therefore it can not displace chlorine from sodium chloride.In the reaction above between lithium bromide and chlorine, 10g of lithium bromide is used. 2LiBr(aq) + Cl2(aq) ? 2LiCl(aq) + Br2How much chlorine will be needed for a complete reaction? Number of Moles LiBr = mass/Mr = 10/87 = 0.115 moles2 LiBr: Cl2 so number of moles Cl2 = 0.115/2 = 0.0575 molesMass Cl2 = n x Mr = 0.0575 x 71 = 4.08gWhat mass of lithium chloride will be produced?Number of moles LiCl = number of moles LiBr = 0.115Mass of LiCl produced = n x Mr = 0.115 x 42 = 4.83gLithium chloride solution conducts electricity because LiCl dissociates to give Li+ and Cl- ions. These ions move through the substance carrying charge and completely the circuit.A solution of lithium chloride is made with 20g of lithium chloride in 100cm3 of water. Calculate its concentration in g/dm3 (triple only – calculate also in mol/dm3).Concentration = mass/ volume = 20/ (100/1000) = 20/0.1 = 200 g/dm3 (cm3 to dm3 divide by 1000)Concentration in g/dm3 → mol/ dm3 divide by Mr. 200 g/dm3 /42 = 4.76 mol/ dm3Fluorine is more reactive than iodine as fluorine atoms are smaller than iodine atoms. This means that electrons are more strongly attracted by the protons of the nucleus of Fluorine forming 1- ions more easily. (Iodine atoms are larger, have more electrons that shield the positive charge of the nucleus so electrons are not as strongly attached meaning iodine is less reactive.Summary problemThis question is about the extraction of copper, an important metal in a number of settings.The bonding of copper. Lattice of positive ions surrounded by delocalised electrons that can move through the whole structure..An alloy of copper is harder than pure copper as in the alloy there are different sized atoms that distort the layers so they aren't able to slide over each other as easily.Copper conducts electricity as the delocalised electrons move through the structure, carrying charge to complete the circuit.Copper is a very good thermal conductor as the delocalised electrons move through the structure, transferring heat energy through the structure.Isotopes are atoms of the same element with the same number of protons but a different number of neutrons.Copper can react with oxygen to form copper oxide (CuO)Copper + Oxygen → Copper Oxide2 Cu (s) + O2(g) → 2 CuO (s)Elements- copper and oxygen. Compound- copper oxide. Reactants- copper and oxygen. Product- copper oxide.If 50g of copper is reacted in an excess of oxygen, what mass of copper oxide will be formed?n Copper = m/Mr = 50g/ 63.5 = 0.787 molesn Copper Oxide = n Copper = 0.787 molesm Copper oxide = n x Mr = 0.787 x (63.5 + 16) = 0.787 x 79.5 = 62.6 g (3 s.f.)Excess oxygen assumes there is so much oxygen preset that it will not run out before the other reactant(s).Copper can be extracted from copper oxide by reacting it with carbon. This is called reduction. Reduction is removal of oxygen or a gain of electrons.Copper Oxide + Carbon → Carbon dioxide + Copper2 CuO(s) + C(s) → CO2(g) + 2 Cu(s)What mass of carbon will be needed to react with 75g of copper oxide?n Copper Oxide = m/Mr = 75/ 79.5 = 0.943 molesn carbon= n Copper Oxide / 2 = 0.943/2 = 0.472 molesm carbon = n x Mr = 0.472 x 12 = 5.66 gCopper can react with some acids to form salts.A weak acid only partially dissociated in solution so there aren’t enough H+ ions to react with copper.A concentrated acid is when an acidic solution has many acid particles present and not very much water present.Copper + hydrochloric acid → copper (II) chloride + hydrogenThe mass of this reaction appears to decrease over time as hydrogen gas is produced which goes into the atmosphere.Cu (s) + 2 HCl (aq) → CuCl2 (aq) + H2 (g)Cu (s) + 2 H+ (aq) → Cu2+ (aq) + H2 (g)Cu (s) → Cu2+ (aq) + 2 e- (aq) and 2 H+ (aq) + 2 e- (aq) → H2 (g)Copper atoms have been oxidised to copper ions. Hydrogen ions have been reduced to hydrogen gas.Part 7: Electrolysis Solid sodium chloride must be melted (molten) or be made into a solution before it can be electrolysed.The electrolyte is the liquid sodium chloride.To electrolyse sodium chloride you need a breaker, two graphite electrodes, two wires, battery, two crocodile clips, sodium chloride solution or solid sodium chloride with a heater.Electrolysis of liquidsIn the electrolysis of zinc chloride, the electrolyte is molten or aqueous zinc chloride.In the electrolysis of each of the molten compounds below, state which elements will be produced:Zinc iodide- zinc and iodineLithium bromide- Lithium and bromineIron fluoride- Iron and fluorineSodium oxide- Sodium and oxygenPotassium chloride- Potassium and chlorineMetals are always produced at the negative electrode (cathode). This would be zinc, lithium, iron, sodium and potassium. Non-metals are produced at the positive electrode (anode). This would be iodine, bromine, fluorine, oxygen and chlorine. Zinc iodide-ZnI2Lithium bromide- LiBrIron fluoride- FeF2 or FeF3Sodium oxide- Na2OPotassium chloride- KClElectrolysis can not be performed on covalent substances as they do not contain any charged particles.Electrolysis can not be performed on metals as it is used to split up compounds and metals are elements.Redox and half equations recapCopy the table below into your exercise book. Add 7 empty plete the tableFormulaPositive ionNegative ionElement formed at cathodeElement formed at anodeHalf equation at cathodeHalf equation at anodeZnCl2Zn2+(l)Cl-(l)Zn(s)Cl2(g)Zn2+(l) + 2e- ? Zn(s)2Cl-(l) ? Cl2(g) + 2e-NaClNa+(l)Cl-(l)Na(s)Cl2(g)Na+(l) + e- → Na (s)2Cl-(l) ? Cl2(g) + 2e-NaBrNa+(l)Br-(l)Na(s)Br2(g)Na+(l) + e- → Na (s)2Br-(l) ? Br2(g) + 2e-KIK+(l)I-(l)K+ (l)O2(g)K+(l) + e- → K (s)2I-(l) ? I2(g) + 2e-CaCl2Ca2+(l)Cl-(l)Ca2+ (aq)Cl2(g)Ca2+(l) + 2 e- → Ca (s)2Cl-(l) ? Cl2(g) + 2e-AlBr3Al3+ (l)Br-(l)Al3+Br2(g)Al3+ (l) + 3 e- → Al (s)2Br-(l) ? Br2(g) + 2e-Na2ONa+(l)O2- (l)Na(s)O2(g)2 Na+(l) + e- → Na (s)2O2- → O2 (g) + 4e-Al2O3Al3+ (l)O2- (l)Al3+O2(g)Al3+ (l) + 3 e- → Al (s)2O2- → O2 (g) + 4e-An ore is a naturally occurring solid material from which a metal or valuable mineral can be extracted profitably.Gold is found naturally in the Earth’s crust as it is very unreactive.Electrolysis is not necessary to extract iron from iron oxide as iron is less reactive than carbon so displacement with carbon can be used.What properties would you expect iron oxide to have? Metal oxides are usually solid and have basic properties.Iron (III) oxide + carbon → iron + carbon dioxideElectrolysis of calcium oxide required to extract pure metal calcium because calcium is more reactive than carbon do cannot be extracted by displacement.2 Fe2O3 + 3 C → 4 Fe + 3 CO2Challenge: The formula of Vanadium (V) Oxide is V2O5. V5+ + 5 e- → V and 2 O2- → O2 → 4e-Electrolysis of aluminium oxideExplain why the formula of aluminium oxide is Al2O3 and not AlO3. Aluminium forms a 3+ ion by losing its three outer shell electrons when it reacts. Oxygen loses 2 electrons when it reacts to form O2- ions. Compounds are neutral, electrons lost from one atom must be gained by another so 6 electrons are lost from 2 aluminum atoms and are transferred to 3 oxygen atoms. They take 2 electrons each.Balance the equation:4 Al + 3 O2 ? 2 Al2O32 aluminium atoms lose 3 electrons each. They are transferred to 3 oxygen atoms who take 2 electrons each.Aluminium oxide has a high melting point because it has a giant lattice structure of strong ionic bonds that require a high amount of energy to break.Aluminium oxide needs to be molten before it can be electrolysed so that the ions are able to move to complete the circuit.In graphite, each carbon atom is connected to three other carbon atoms leaving one electron delocalised per atom. These electrons can move through the structure, carrying charge to complete the circuit and conduct electricity.The anodes must be regularly replaced as the oxygen produced there reacts with the graphite it is made from producing carbon dioxide.2 C (s) + O2 (g) → 2 CO (g) Electrolysis is an endothermic change. Electrolysis is endothermic as energy is required for the reaction to take place.Electrolysis of solutions For each of the below, state which elements are formed at the anode and at the cathodeSolutionProduct at the anodeProduct at the cathodeCopper SulfateOxygenCopperSilver NitrateOxygenSilverTin ChlorideChlorineHydrogenZinc FluorideFluorineHydrogenZinc SulfateOxygenHydrogenCalcium NitrateOxygenHydrogenPotassium ChlorideChlorineHydrogenPotassium can only be extracted from potassium nitrate if it is molten, not if it is dissolved as potassium is more reactive than hydrogen so hydrogen gas would be discharged at the cathode.Sodium chloride is dissolved in water. Na+ and Cl- and present when sodium chloride dissolves.Na+ and Cl- and H+ and OH- (from the water) are present when sodium chloride dissolves.Chlorine gas will be produced at the anode.Hydrogen gas will be produced at the cathode.Half equations-Anode: 2Cl- (aq) → Cl2(g) + 2e-(aq)Cathode: 2 H+ (aq) + 2e-(aq) → H2(g)The process is repeated but with copper (II) sulphate. Give half equations for the reactions at the anode and the cathodeAnode: 4OH- (aq)→ O2(g) + 2H2O(l) + 4 e-Cathode: Cu2+ (aq) + 2e- → Cu (s)GCSE QuestionsQ1.Iodine has a higher boiling point than chlorine as iodines molecules are larger so it has greater intermolecular forces and requires more energy to boil.Boiling point of bromine- anywhere from -34 to 183. Most likely somewhere in the middle around 75 degrees celsius. Google says 59 degrees celsius.Ionic equation for the reaction of chlorine with potassium iodide: Cl2(aq) + 2 I-(aq) → 2Cl-(aq) + I2(aq)Potassium iodide solution conducts electricity as it is an ionic compound and the ions are able to move when it is aqueous.The products of electrolysing potassium iodide are hydrogen and iodine.Q2.(a)?????When zinc chloride is molten, it will conduct ELECTRICITY .(1)(b)?????Zinc ions move towards the negative electrode where they gain electrons to produce zinc.(i)??????Name the product formed at the positive electrode. CHLORINE (GAS) (1)(ii)?????Zinc ions move towards the negative electrode as they are positively charged (Zn2+). Opposite charges attract.. (2)(iii)????Zinc is reduced when is gains electrons (1)(c)?????Zinc is mixed with copper to make an alloy.(i)?????The alloy is harder than pure zinc as the alloy has different sized atoms that distort the layers meaning they don’t slide over each other. (2)Q3. This question is about magnesium and magnesium chloride.(a)?????When a magnesium atom reacts with two chlorine atoms, the magnesium atoms loses two electrons to form magnesium ions (Mg2+). Two chlorine atoms gain one of these electrons each to form chloride ions (Cl?).(b)?????(i)??????Magnesium chloride must be molten or dissolved in water to be electrolysed so that the ions are free to move. In a solid they are in a fixed position so could not carry charge to complete the circuit.. (2)(ii)?????The magnesium ions are attracted to the cathode (negative electrode) where they accept two electrons to form solid magnesium. (3)(iii)????Chlorine gas produced at the negative electrode. (1)(iv)?????Magnesium is not produced at the negative electrode in Experiment 2 as hydrogen ions are present. As hydrogen is less reactive than Magnesium it will be produced at the cathode (1)(v)????2 Cl????????→???????Cl2???????+???????2 e- (1)(c)?????Metals can be bent and shaped as their atoms are arranged in layers that can slide over each other.. (2) ................
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