AP Chemistry



Chemistry 5: Stoichiometry—General Name __________________________A.Chemical Reactions (3.1-3.4, 3.7)1.chemical equation93408523685500a.coefficients and subscripts1 H2Osubscript refers to # ofatom that precedes it2 H2Ocoefficient refers to # ofmolecules that followb.reactants and products1.one directional reaction: reactants ? products2.equilibrium reaction: "reactants" ? "products"c.conservation of atoms (mass)—Dalton's Theory CH4(g) + 2 O2(g) ? CO2(g) + 2 H2O(g) 1 C 4 O 1 C 4 H 4 H 2 O 2 O (16 g) + (64 g) = (44 g) + (36 g)2.types of chemical reactionsa.non-aqueous reactionsTypeExamplecombination2 Mg(s) + O2(g) ? 2 MgO(s)decompositionCaCO3(s) ? CaO(s) + CO2(g)combustionC5H12(l) + 8 O2(g) ? 5 CO2(g) + 6 H2O(g)b.aqueous reactions discussed next unit3.calculations based on balanced chemical equationsa.coefficients represent moles of formula unitsb. model calculations_ g A x 1 mol A x (#) mol B x (MM) g B = _ g B (MM) g A (#) mol A 1 mol B_ L A x (M) mol A x (#) mol B x __1 L B__ = _ L B 1 L A (#) mol A (M) mol Bc.for gas reactants and products: moles and volume are proportional to coefficients (22.4 L/mol at STP)4.limiting reactant and theoretical yield a.definitions1.limiting reactant—reactant consumed first 2.theoretical yield—maximum product made Before reactionAfter reaction10 H2 + 7 O2 10 H2O + 2 O2 H2 is limiting reactant10 H2O is theoretical yield2 O2 is excess reactantb.procedurecalculate moles of each reactant availablecalculate moles of one product based on moles of each reactant ? smallest = theoretical yield use theoretical yield for remaining calculationsexcess reactant = mole present – moles usedpercent yield = 100(actual yield/theoretical yield)B.Gravimetric Analysis (3.5)1.mass percent from formulamoles ? mass for each element (MM x subscript)add masses to get total massmass % = 100(mass part/total mass)2.empirical formulaconvert g (or %) ? molesdivide each mole value by smallestmultiple by factor to make all whole numberswhole numbers become subscripts"burning" carbon compounds yield CO2 and H2OCxHyOz(g) + _ O2(g) ? X CO2(g) + Y/2 H2O(g)mole C = mole CO2 = mCO2/44mole H = 2 mole H2O = mH2O/9mole O = (mCxHyOz – mC – mH)/163.molecular formula (given MM)MM/empirical formula mass = constantmultiple each subscript in empirical formula by constant = molecular formulaExperiments1.Percent Yield of CO2 Gas Lab (Wear Goggles)—Measure the amount of CO2(g) produced in a reaction and compare it to the theoretical yield.Mass 0.6 g NaHCO3 and record its mass (m) to the nearest 0.001 g. Add the NaHCO3 to the flask and ? fill the pipet with 6 M HCl. Assemble the gas generating apparatus. React all of the NaHCO3. Measure the volume of water remaining in the bottle (V1) and the capacity of the gas collecting bottle (V2). Measure the water temperature (T). Record the room pressure (Plab) and water vapor pressure (PH2O).a.Record the collected data.m(g)V1(mL)V2(mL)T(oC)Plab(torr)PH2O(torr)b.Calculate the following for the CO2 gas.P (atm)V (L)T (K)c.Determine the moles of CO2 produced based on the mass of NaHCO3 reacted (limiting reactant).HCO3- + H+ ? H2O + CO2(g)d.Determine the theoretical volume of CO2 produced at lab conditions based on moles of CO2 reacted. e.Determine the % yield of CO2 gas from the actual volume collected and the theoretical volume.f.Suggest a possible reason why the yield was less than 100 %.2.Analysis of a Hydrate Lab (Wear Goggles)—Determine the moles of water in a hydrate by massing the hydrate and anhydrous after removing the water by heating.Mass a clean, dry 150-mL beaker (m1). Add a spoonful of hydrated CaSO4 to the beaker and mass the beaker + hydrate (m2). Place the beaker on a hot plate (set at 7) for 15 minutes. Place the beaker on the lab table to cool. Mass the beaker + anhydrous (m3). Return the beaker to the hot plate for an additional 5 minutes of heating, and then mass it again (m3). If the beaker loses additional mass, then repeat the heating and massing until you get two masses that are within 0.002 g of each other. a.Record the collected data. m1 (g)m2 (g)m3 (g)b.Calculate the masses, percentages and moles.HydrateAnhydrousWaterMassMass %100 %Molesc.The formula for hydrated calcium sulfate is CaSO4?X H2O. Determine the value of X in the formula.Practice ProblemsA.Chemical Reactions1.Balance the chemical equations.__NH3(g) + __O2(g) ? __NO(g) + __H2O(g)__(NH4)2CO3(s) ? __NH3(g) + __CO2(g) + __H2O(g)__Fe2O3(s) + __H2(g) ? __Fe(s) + __H2O(l) __N2H4(g) + __H2O2(l) ? __N2(g) + __H2O(l)plete and balance the chemical bination__Na(s) + __O2(g) ? __Na2O(s)__K(s) + __I2(s) ? __KI(s)__Zn(s) + __O2(g) ? __ZnO(s)Decomposition__BeC2O4?3 H2O(s) ? __BeC2O4(s) + __H2O(g)__CaCO3(s) ? __CaO + __CO2(g)Combustion__CH4(g) + __C2H5OH(l) + __HC3H5O2(l) + __C3H8(g) +__C3H7OH(l) + 3.CH4(g) + 2 O2(g) ? CO2(g) + 2 H2O(l). How many grams of O2 react with 3.87 g of CH4?4.Fe2O3(s) + 3 H2(g) ? 2 Fe(s) + 3 H2O(l) How many liters of H2 (d = 0.816 g/L) are needed to produce 10.0 g Fe?5.2 K(s) + I2(s) ? 2 KI(s)How many grams of KI are produced when 6.03 g K react?6.MnO2(s) + 2 Cl-(aq) + 4 H+(aq) ? Mn2+(aq) + Cl2(g) + 2 H2O(l) How many grams of MnO2 are required to produce 1.20 L of Cl2 gas (d = 1.83 g/L)?7.N2(g) + 3 H2(g) ? 2 NH3(g)1.26 g of N2 reacts with 0.300 g H2 forming 0.874 g of NH3. a.What is the limiting reactant?b.What is the theoretical yield?c.What is the percent yield?8.N2H4(g) + 2 H2O2(l) ? N2(g) + 4 H2O(l)2.69 g of N2H4 reacts with 3.14 g of H2O2. Determine thea.limiting reactant. b.mass of N2, H2O, N2H4 and H2O2 after the reaction.9.2 NH3(g) + H2S(g) ? (NH4)2S(s)How many grams of NH3, H2S and (NH4)2S are present after 6.84 g of NH3 reacts with 4.13 g of H2S?a.What is the Theoretical yield of (NH4)2S(s)? b.What is the mass of (NH4)2S(s) produced? c.What is the mass of excess reactant remaining?10. 4 Fe(s) + 3 O2(g) ? 2 Fe2O3(s)75.0 g of Fe is mixed with 11.5 L of O2 (d = 3.48 g/L). Calculate the mass of Fe2O3 produced.B.Gravimetric Analysis11.What are the mass percentages of each element in KMnO4?12.a.Calculate the mass percent of each element in Al2O3. b.How much Al can be obtained from 32.0 g of Al2O3? 13.A mixture of sulfur and iron has a mass of 13.6841 g. A magnet separates the iron, which has a mass of 0.286 g. a.What is the mass of the sulfur?b.What is the mass percent of sulfur?14. Concentrated nitric acid (HNO3) is 70.8 % by mass and has a density of 1.42 g/mL. a.How many moles of HNO3 are in 100 g of solution?b.What is the volume of 100 g of solution?c.What is the molarity of the solution?15.What is the formula of compound that is 31.9% K, 29.0% Cl, and 39.2% O?16.A compound is made up of C, Cl, and O. It is 12.1% C and 70.9% Cl by mass. What is its empirical formula?17.A sample of compound contains 78.2 g K and 32.1 g S. What is its formula?18.A compound is made up of 16.9 g of sodium, 11.8 g of sulfur, and 23.6 g of oxygen. What is its empirical formula?19.1.08 g of a compound containing C and S burns in air and generates 0.627 g CO2. What is its empirical formula?20.1.500 g of compound containing only C, H and O is burned in excess of oxygen. 1.433 g of CO2 and 0.582 g of H2O are produced. What is the empirical formula?21.What is the molecular formula of a compound that has the empirical formula, CH3O, and a molecular mass of 62.0 g/mol?22.Benzene has the empirical formula of CH and a molecular mass of 78.0 g/mol. What is its molecular formula? 23. A compound contains the elements C, H, N, and O. a.When a 1.2359 g sample is burned in excess oxygen, 2.241 g of CO2 and 0.5781 g H2O are formed. (1)Determine the mass of C and H in the sample.(2)The mass percent of N is found to be 28.84%. Determine the mass of N in the 1.2359 g sample.(3)Determine the mass of O in the 1.2359 g sample.b.Determine the empirical formula of the compound.c.The molecular mass of the compound is approximately 300 g/mol. Determine the molecular formula.24. 75.0 g of Fe reacts with 11.5 L of O2 (d = 3.48 g/L) according to the reaction: 4 Fe(s) + 3 O2(g) ? 2 Fe2O3(s)a.Calculate the initial moles of Fe(s) and O2(g). b.Determine the limiting reactant with calculations.c.Calculate the mass of Fe2O3(s) produced.Practice Multiple ChoiceBriefly explain why the answer is correct in the space provided.1._ Fe2O3 + _ CO ? _ Fe + _ CO2When the equation is balanced and reduced to lowest terms, the coefficient for CO2 is (A)1(B)2(C)3(D)42.1 CH3CH2COOH + _ O2 ? _ CO2 + _ H2OHow many moles of O2 are required to oxidize 1 mole of CH3CH2COOH according to the reaction above?(A)2(B)5/2(C)3(D)7/23. C3H8 burns in excess oxygen gas. What is the coefficient for O2 when the equation is balanced with lowest terms?(A)4(B)5(C)7(D)104. CaCO3 + 2 HCl ? CaCl2 + CO2 + H2OWhat is the mass percent of CaCO3 in a 1.25-g rock that generate 0.44 g of CO2 when reacted with HCl? (A)35 % (B)44 %(C)67 %(D)80 %5. 8.0 mol of F2 and 1.7 mol of Xe are mixed. When all of the Xe reacted, 4.6 mol of F2 remain. What is the formula? (A)XeF(B)XeF3(C)XeF4 (D)XeF6 6.What mass of Ca(NO3)2 contains 24 g of oxygen atoms?(A) 164 g(B) 96 g(C) 62 g(D) 41 pounds contain 38 g, 57 g, 76 g, and 114 g of element Q per mole compound. A possible atomic mass of Q is(A) 13(B) 19(C)28(D) 388.What is the percent nitrogen by mass in N2O3?(A) 18 %(B) 22 %(C) 36 %(D) 45 %9.Which formula is 54 % water by mass?(A)CaCO3 ? 10 H2O(B) CaCO3 ? 6 H2O(C) CaCO3 ? 2 H2O(D) CaCO3 ? H2O10.Which formula forms 88 g of carbon dioxide and 27 g of water when burned in excess oxygen?(A) CH4(B) C2H2(C) C4H3(D) C4H611.How many moles of H2O are produced when 0.56 g of C2H4 (MM = 28 g) is burned in excess oxygen?(A) 0.04(B) 0.06(C) 0.08(D) 0.4 12. What is the simplest formula of a compound that contains 1.10 mol of K, 0.55 mol of Te, and 1.65 mol of O? (A)KTeO (B)KTe2O (C)K2TeO3 (D)K2TeO6 13. In which compound is the mass ratio of chromium to oxygen closest to 1.6 to 1.0? (A)CrO3(B)CrO2(C)CrO(D)Cr2O14. In which is the mass percent of magnesium closest to 60 %. (A)MgO(B)MgS(C)MgF2 (D)Mg3N2 15. A student obtained a percent water in a hydrate that was too small. Which is the most likely explanation for this? (A)Hydrate spattered out of the crucible during heating(B)The anhydrous absorbed moisture after heating. (C)The amount of hydrate sample used was too small. (D)The amount of hydrate sample used was too large.16. 2 N2H4 + N2O4 ? 3 N2 + 4 H2OWhat mass of water can be produced when 8.0 g of N2H4 (MM = 32 g) and 9.2 g of N2O4 (MM = 92 g) react? (A)9.0 g(B)18 g(C)36 g(D)7.2 g17. What mass of Au is produced when 0.0500 mol of Au2S3 is reduced completely with excess H2? (A)9.85 g (B)19.7 g(C)24.5 g(D)39.4 gPractice Free plete and balance the chemical equation.__C3H10(g) + __O2(g) ? __CO2(g) + __H2O(g)__C4H9OH(l) + __O2(g) ? __CO2(g) + __H2O(g)__NH4NO3(s) ? __N2O(g) + __H2O(g)__Al(s) + __O2(g) ? __Al2O3(s)__Li(s) + __N2(g) ? __Li3N(s)2.6 Li(s) + N2(g) ? 2 Li3N(s)A mixture of 5.00 g of Li and N2 react.a.What is the limiting reactant?b.How much excess reactant is there?c.How many grams of Li3N are formed?d.If the percent yield is 88.5%, how many grams of Li3N are produced? 3.Nicotine, a component of tobacco, is composed of C, H and N. A 5.250-g sample of nicotine was combusted, producing 14.242 g of CO2 and 4.083 g of H2O. What is the empirical formula of nicotine?4. Answer the following questions about acetylsalicylic acid. a.What is mass percent of acetylsalicylic acid in a 2.00-g tablet that contains 0.325 g acetylsalicylic acid?b.Acetylsalicylic acid contains H, C and O. Combustion of 3.00 g yields 1.20 g H2O and 3.72 L of CO2 at 50oC and 1.07 atm, calculate the mass of each element.c.Determine the empirical formula of acetylsalicylic acid.5.Bismuth (Bi) reacts with fluorine to form BiF3.a.Calculate the mass percent of Bi and F in the compound.b.Calculate the mass of fluorine required to form 16.5 g of compound.c.Write a balanced equation for the reaction.d.How many moles of F2 are required to react with 0.240 mol Bi?e.How many grams of F2 are required to react with 1.60 g Bi?f.If 5.00 g of Bi react with 2.00 g F2, what is the limiting reactant?g.What is the theoretical yield of BiF3 when 5.00 g Bi and 2.00 g F2 react?h.When BiF3 reacts with water, one of the products is HF. The other product is a compound that contains 85.65 % Bi, 6.56 % O, and 7.79 % F. What is the simplest formula of this compound?i.Write a balanced equation for the reaction between BiF3 and H2O.6. Answer the following questions that relate to the analysis of chemical compounds. a.A compound containing the elements C, H, N, and 0 is analyzed. When a 1.2359 g sample is burned in excess oxygen, 2.241 g of CO2 (g) is formed. The combustion analysis also showed that the sample contained 0.0648 g of H. (1)Determine the mass, in grams, of C in the 1.2359 g sample of the compound. (2)When the compound is analyzed for N content only, the mass percent of N is found to be 28.84 percent. Determine the mass, in grams, of N in the original 1.2359 g sample of the compound. (3)Determine the mass, in grams, of 0 in the original 1.2359 g sample of the compound.(4)Determine the empirical formula of the compound. b.A different compound, which has the empirical formula CH2Br, has a vapor density of 6.00 g?L-1 at 375 K and 0.983 atm. Using these data, determine the following. (1)The molar mass of the compound (2)The molecular formula of the compound ................
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