Page 170 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 10] STOICHIOMETRY 159

Since they react in a 1:1 mole ratio, BaCl 2 is obviously in limiting quantity.
2 mol NaCl 58.5 g NaCl

0.0909 mol BaCl 2 = 10.6 g NaCl
1 mol BaCl 2 1 mol NaCl
10.59. If 4.00 g Cu is treated with 15.0 g AgNO 3 , how many grams of Ag metal can be prepared?

2 AgNO + Cu −→ 2Ag + Cu(NO 3 ) 2
3
1 mol Cu
Ans. 4.00gCu = 0.0630 mol Cu present
63.5gCu
1 mol AgNO

15.0 g AgNO 3 = 0.0882 mol AgNO present
3 3
170 g AgNO 3
0.0882 mol AgNO 3
= 0.0441 mol AgNO 3
2
0.0630 mol Cu
= 0.0630 mol Cu
1
AgNO 3 is present in limiting quantity, and we base the calculation on the AgNO 3 present.

1 mol Ag 108gAg
0.0882 mol AgNO 3 = 9.53gAg
1 mol AgNO 3 1 mol Ag
10.60. What, if any, is the difference between the following three exam questions?
How much NaCl is produced by treating 10.00 g NaOH with excess HCl?
How much NaCl is produced by treating 10.00 g NaOH with sufficient HCl?
How much NaCl is produced by treating 10.00 g NaOH with HCl?
Ans. There is no difference.

10.61. What mass of water can be heated from 20.0 C to 45.0 C by the combustion of 4.00 g of carbon to carbon dioxide?
◦
◦
C + O 2 −→ CO 2 + 393 kJ

Ans. 4.00gC 1 mol C 393 kJ = 131 kJ produced
12.0gC 1 mol C
Hence 131 kJ = 131 000 J is added to the water.
heat 131 000 J
m = = = 1250 g = 1.25 kg
(c)( t) [4.184 J/(g· C)](25.0 C)
◦
◦
Note that the heat given off by the reaction is absorbed by the water.
10.62. Some pairs of substances undergo different reactions if one or the other is in excess. For example,
With excess O 2 : C + O 2 −→ CO 2
With excess C : 2 C + O 2 −→ 2CO

Does this fact make the limiting-quantities calculations in the text untrue for such pairs?
Ans. No. The principles are true for each possible reaction.

10.63. A manufacturer produces impure ammonium phosphate as a fertilizer. Treatment of 56.0 g of product with excess
NaOH produces 18.3 g NH 3 . What percentage of the fertilizer is pure (NH 4 ) 3 PO 4 ? (Assume that any impurity
contains no ammonium compound.)
Ans. 3 NaOH + (NH 4 ) 3 PO 4 −→ 3NH 3 + 3H 2 O + Na 3 PO 4

1 mol NH 3 1 mol (NH 4 ) 3 PO 4 149 g (NH 4 ) 3 PO 4
18.3gNH 3 = 53.5g (NH 4 ) 3 PO 4
17.0gNH 3 mol NH 3 1 mol (NH 4 ) 3 PO 4
3

53.5g (NH 4 ) 3 PO 4
× 100% = 95.5% pure
56.0 g product