Page 169 - Theory and Problems of BEGINNING CHEMISTRY

158 STOICHIOMETRY [CHAP. 10

Ans.
1 mol C 6 H 12 1 mol C 6 H 11 Br 163gC H 11 Br
6
5.00gC H 12 = 9.68gC H 11 Br
6
6
84.2gC H 12 1 mol C 6 H 12 1 mol C 6 H 11 Br
6
5.00gC H 11 Br obtained × 100% = 51.7% yield
6
9.68gC H 11 Br possible
6
10.53. In a certain experiment, 225 g of ethane, C 2 H 6 , is burned. (a) Write a balanced chemical equation for the combustion
of ethane to produce CO 2 and water. (b) Determine the number of moles of ethane in 225 g of ethane. (c) Determine
the number of moles of CO 2 produced by the combustion of that number of moles of ethane. (d) Determine the mass
of CO 2 that can be produced by the combustion of 225 g of ethane.
Ans. (a)2 C 2 H 6 + 7O 2 −→ 4CO 2 + 6H 2 O

1 mol C 2 H 6
(b) 225gC 2 H 6 = 7.48 mol C 2 H 6
30.1gC H 6
2

4 mol CO 2
(c) 7.48 mol C 2 H 6 = 15.0 mol CO 2
2 mol C 2 H 6
44.0gCO 2

(d ) 15.0 mol CO 2 = 660gCO 2
1 mol CO 2
10.54. Determine the number of kilograms of SO 3 produced by treating excess SO 2 with 50.0 g of oxygen.
Ans. 2SO 2 + O 2 −→ 2SO 3

1 mol O 2 2 mol SO 3 80.0gSO 3
50.0gO 2 = 250gSO = 0.250 kg SO 3
3
32.0gO 2 1 mol O 2 1 mol SO 3
10.55. How many grams of K 2 CO 3 will be produced by thermal decomposition of 4.00 g KHCO 3 ?
heat
Ans. 2 KHCO 3 −→ K 2 CO 3 + CO 2 + H 2 O

1 mol KHCO 3 1 mol K 2 CO 3 138gK CO 3
2
4.00 g KHCO 3 = 2.76gK 2 CO 3
100 g KHCO 3 2 mol KHCO 3 1 mol K 2 CO 3
−
10.56. What is the difference in mass between reactions involving 2.25 g Cl and 2.25 g NaCl?
Ans. The former might be part of any ionic chloride, but more importantly, the former has a greater mass of
−
chlorine, because 2.25 g NaCl has less than 2.25 g Cl .
10.57. How many moles of H 2 can be prepared by treating 5.00 g Na with 2.80 g H 2 O? Caution: This reaction is very
energetic and can even cause an explosion.
Ans. 2Na + 2H 2 O −→ 2 NaOH + H 2

1 mol Na
5.00gNa = 0.217 mol Na
23.0gNa

1 mol H 2 O
2.80gH O = 0.156 mol H 2 O
2
18.0gH O
2
Since they react in a 1 : 1 mole ratio, H 2 O is obviously in limiting quantity.

1 mol H 2
0.156 mol H 2 O = 0.0780 mol H 2
2 mol H 2 O
10.58. How many grams of NaCl can be prepared by treating 18.9 g BaCl 2 with 14.8 g Na 2 SO 4 ?
Ans. BaCl 2 + Na 2 SO 4 −→ BaSO 4 + 2 NaCl

1 mol BaCl 2
18.9 g BaCl 2 = 0.0909 mol BaCl 2
208 g BaCl
2

1 mol Na 2 SO 4
14.8gNa SO 4 = 0.104 mol Na 2 SO 4
2
142gNa SO 4
2