- E—The balanced chemical equation is:
2 Au 2 O 3 →4 Au +3 O 2(221 g Au 2 O 3 )(l mol Au 2 O 3 /442 g Au 2 O 3 )(3 mol O 2 /2 mol Au 2 O 3 )=0.750 mol O 2- D—The balanced equation is:
4 C 4 H 11 N(l) +27 O 2 (g) →16 CO 2 (g) +22 H 2 O(1) +2 N 2 (g)- E—The KMnO 4 is the limiting reagent. Each mole of KMnO 4 will produce a mole of
MnSO 4. - D—The balanced equation is:
2 KClO 3 →2 KCl +3 O 2
(4.0 mol KClO 3 )(3 mol O 2 /2 mol KCIO 3 ) =6.0 mol O 2- E—The balanced equation is:
2 C 8 H 18 +25 O 2 →16 CO 2 +18 H 2 O
(1.00 mol C 8 H 18 )(25 mol O 2 /2 mol C 8 H 18 ) =12.5 mol O 2- E—(0.200 mol Ca)(l mol H 2 /l mol Ca)(22.4 L at STP/l mol H 2 )=4.48 L
- C—There are 5.00 × 10 −^3 mol of CrO 42 −and an equal number of mol of SnO 22 −. Thus
SnO 22 −is the limiting reactant (larger coefficient in the balanced reaction).
(5.00 × 10 −^3 mol SnO 22 −)(2 mol OH−/3 mol SnO 22 −) =3.33 × 10 −^3 mol OH−- D—The volume of water is irrelevant.
0.20 mol of KBr will require 0.10 mol of Pb(NO 3 ) 2
0.20 mol of MgBr 2 will require 0.20 mol of Pb(NO 3 ) 2
Total the two yields.Free-Response Questions
Answer the following questions. You have 15 minutes, and you may use a calculator and
the tables in the back of the book.A sample of a monoprotic acid was analyzed. The sample contained 40.0% C and 6.71% H.
The remainder of the sample was oxygen.a. Determine the empirical formula of the acid.b. A 0.2720-g sample of the acid was titrated with standard NaOH solution. Determine
the molecular weight of the acid if the sample required 45.00 mL of 0.1000 M NaOH
for the titration.c. A second sample was placed in a flask. The flask was placed in a hot water bath until
the sample vaporized. It was found that 1.18 g of vapor occupied 300.0 mL at 100°C
and 1.00 atmospheres. Determine the molecular weight of the acid.d. Using your answer from part a, determine the molecular formula for part band for part c.e. Account for any differences in the molecular formulas determined in part d.Stoichiometry 99