Chemistry 199



Chemistry 121 Oregon State University

Worksheet 8 Notes Dr. Richard L Nafshun

1. Given the following reaction: C3H8 + 5 O2 ⎝ 3 CO2 + 4 H2O

Assume that 10.00 g of propane react with excess oxygen. Calculate:

(A) the number of moles of propane that react

10.00 g C3H8 [pic] = 0.2267 mol propane

(B) the number of molecules of propane that react;

0.2267 mol propane [pic] = 1.365 x 1023 propane molecules

(C) the number of moles of carbon dioxide and water that are formed;

0.2267 mol propane [pic] = 0.6801 mol CO2

0.2267 mol propane [pic] = 0.9068 mol H2O

(D) the number of grams of carbon dioxide and water that are formed;

0.6801 mol CO2 [pic] = 29.93 g CO2

0.9068 mol H2O [pic] = 16.34 g H2O

(E) the number of molecules of carbon dioxide and water that are formed;

0.6801 mol CO2 [pic] = 4.096 x 1023 CO2 molecules

0.2267 mol propane [pic] = 0.9068 mol H2O

0.9068 mol H2O [pic] = 5.461 x 1023 H2O molecules

(F) the number of moles, grams and molecules of oxygen that are used up.

0.2267 mol propane [pic] = 1.134 mol O2

1.134 mol O2 [pic] = 6.826 x 1023 O2 molecules

1.134 mol O2 [pic] = 36.29 g O2

(G) a student collects 23.50 grams of CO2. What is the percent yield?

Percent Yield = [pic] = [pic](100%) = 78.52%

2. Calculate the molarity of the following aqueous solutions.

(A) 6.00 mol HCl in 2.50 L of solution.

M = [pic] = [pic] = 2.40 M

(B) 45.00 grams NaOH in 10.00 L of solution.

M = [pic] = [pic] = 0.1125 M

3. How many grams of NaOH are required to prepare 50.0 mL of 0.125 molar NaOH?

M = [pic] = [pic] = 0.125 M

X = 0.251 g NaOH

4. (A) What is the mass percent composition of Na, S and O in the compound Na2SO4?

The molar mass of Na2SO4 is 142.06 g/mol

The mass of sodium in Na2SO4 is 2 x 23.00 g/mol = 46.00 g/mol

The mass percent of sodium in Na2SO4 is the part "over the whole."

Mass percent of sodium in Na2SO4 = [pic](100%) = 32.38%

The mass of sulfur in Na2SO4 is 1 x 32.06 g/mol = 46.00 g/mol

The mass percent of sulfur in Na2SO4 is the part "over the whole."

Mass percent of sulfur in Na2SO4 = [pic](100%) = 22.57%

The mass of oxygen in Na2SO4 is 4 x 16.00 g/mol = 64.00 g/mol

The mass percent of oxygen in Na2SO4 is the part "over the whole."

Mass percent of oxygen in Na2SO4 = [pic](100%) = 45.05%

(B) What is the mass percent composition of Ca, N and O in the compound Ca(NO3)2?

The molar mass of Ca(NO3)2 is 164.10 g/mol

The mass of calcium in Ca(NO3)2 is 1 x 40.08 g/mol = 40.08 g/mol

The mass percent of calcium in Ca(NO3)2 is the part "over the whole."

Mass percent of calcium in Ca(NO3)2 = [pic](100%) = 24.42%

The mass of nitrogen in Ca(NO3)2 is 2 x 14.01 g/mol = 28.02 g/mol

The mass percent of nitrogen in Ca(NO3)2 is the part "over the whole."

Mass percent of nitrogen in Ca(NO3)2 = [pic](100%) = 17.07%

The mass of oxygen in Ca(NO3)2 is 6 x 16.00 g/mol = 96.00 g/mol

The mass percent of oxygen in Ca(NO3)2 is the part "over the whole."

Mass percent of oxygen in Ca(NO3)2 = [pic](100%) = 58.50%

5. Given the following reaction: C3H8 + 5 O2 ⎝ 3 CO2 + 4 H2O

Assume that 88.22 g of propane react with excess oxygen. Calculate:

(A) the number of moles of propane that react

88.22 g C3H8 [pic] = 2.000 mol propane

(B) the number of molecules of propane that react;

2.000 mol propane [pic] = 1.204 x 1024 propane molecules

(C) the number of moles of carbon dioxide and water that are formed;

2.000 mol propane [pic] = 6.000 mol CO2

(D) the number of grams of carbon dioxide and water that are formed;

6.000 mol CO2 [pic] = 264.1 g CO2

(E) the number of molecules of carbon dioxide and water that are formed;

6.000 mol CO2 [pic] = 3.612 x 1024 CO2 molecules

2.000 mol propane [pic] = 8.000 mol H2O

8.000 mol H2O [pic] = 4.816 x 1024 H2O molecules

(F) the number of moles, grams and molecules of oxygen that are used up.

2.000 mol propane [pic] = 10.00 mol O2

10.00 mol O2 [pic] = 6.02 x 1024 O2 molecules

10.00 mol O2 [pic] = 320.0 g O2

(G) a student collects 55.90 grams of CO2. What is the percent yield?

Percent Yield = [pic] = [pic](100%) = 21.17 %

6. Balance: C11H24 + O2 ⎝ CO2 + H2O

1 C11H24 + 17 O2 ⎝ 11 CO2 + 12 H2O

How many moles of water are produced if four moles of C11H24 are consumed?

48 (twelve moles of water are produced for every mole of C11H24 consumed.)

4 moles C11H24 [pic] = 48 mol H2O

7. Balance: Li + O2 ⎝ Li2O

4 Li + O2 ⎝ 2 Li2O

How many moles of Li2O are produced if eight moles of Li are consumed?

4 (2 moles of Li2O are produced for every 4 moles of Li consumed.)

8 mol Li [pic] = 4 mol Li2O

8. Determine the mass percent composition of lithium sulfate.

Li2SO4 = 109.94 g/mol

% Li = (2 x 6.941 g/mol)/109.94 g/mol = 12.63 %

% S = (1 x 32.06 g/mol)/109.94 g/mol = 29.16 %

% O = (4 x 16.00 g/mol)/109.94 g/mol = 58.21 %

_______

100.00 %

9. 20.5 g NaCl are dissolved in 500.0 mL of total solution. Calc M.

NaCl = 58.45 g/mol

M = grams/L = (20.5 g/58.45 g/mol) / 0.5000 L = 0.702 M or mol/L

10. 2.50 L of 2.00 M NaF solution is diluted to 4.00 L what is the molarity of the resulting solution?

MBeforeVBefore = MAfterVAfter

(2.00 M)(2.50 L) = (MAfter)(4.00 L)

MAfter = 1.25 M

Abbreviated Solubility Rules (see Table 4.1—Page 128 when working ChemSkill Builder):

Rule 1: All nitrates, group 1A metal salts and ammonium salts are soluble.

Rule 2: All carbonates, hydroxides, phosphates and sulfides are insoluble.

Rule 3: Rule 1 always takes precedent.  Example: NaOH is soluble.

1. A student places 15.00 grams of solid sodium sulfate into a 500.00-mL volumetric flask and fills to the mark with water. Draw a picture of the ions in solution (similar to Figure 4.2—Page 125; but rather than label the ions as (+) and (-), label the ions with the chemical formula. Calculate the number of sodium ions present. Calculate the number of sulfate ions present. Write a balanced equation for the dissociation of the solid ionic compound.

Na2SO4 (s) → 2 Na+ (aq) + SO42- (aq)

Note that two sodium ions and one sulfate polyatomic ion are generated for each sodium sulfate unit that dissociates in water.

15.00 grams Na2SO4 [pic]0.1056 mol Na2SO4

0.1056 mol Na2SO4 [pic]6.359 x 1022 Na2SO4 units

6.359x 1022 Na2SO4 units [pic]1.272 x 1023 Na+ ions

6.359 x 1022 Na2SO4 units [pic]6.359 x 1022 SO42- ions

2. List three strong acids and show how they exist in water (in other words, show a reaction for each that shows them dissociate).

Hydrochloric acid HCl → H+ + Cl-

(because HCl is a strong acid, HCl dissociates 100%).

Nitric acid HNO3 → H+ + NO3-

(because HNO3 is a strong acid, HNO3 dissociates 100%).

Sulfuric acid H2SO4 → H+ + HSO4-

(because H2SO4 is a strong acid, H2SO4 dissociates 100%).

3. Draw the structure of propanoic acid (CH3CH2COOH) and carefully show how it exists in water (in other words, show a reaction that shows it dissociate).

The above reaction does not go 100%. Propanoic acid dissociates less than 2%. So, in water, there is a considerable amount of CH3CH2COOH and very little CH3CH2COO- and H+.

Why does propanoic acid dissociate less than 2%? Because it is a weak acid (it is a carboxylic acid; it contains the –COOH group). Weak acids dissociate less than 100% (usually less than 2%).

The double arrows shown (one forward and one backward) indicate that at any given moment there are reactant (CH3CH2COOH) and products (CH3CH2COO- and H+) present. As a matter of fact, the forward and backward arrows indicate that the system is not static, but dynamic. At any given moment reactant is changing to products and products are changing into reactant (species are changing "positions.")

Warning! Because of cellular phones the following analogy is outdated and is therefore disclosing the age of chemistry faculty.

Think of it like two pay phones in an entire dorm. At any given moment two residents are on the phones (these residents are the products), but 198 residents are not on the phones (these residents are the reactants). When one resident concedes the phone, another resident starts to use the phone. So, there are always two residents on the phones and 198 residents not on the phones, but the residents are changing places (this is dynamic—ever changing). The same two residents are not always on the phones (this would be considered static—fixed).

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