Chapter 9

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 104
- Chapter 9 -
(b)
2 NaOH + H 2 SO 4 ¡ Na 2 SO 4 + 2 H 2 O
10.0 g 10.0 g
Choose one of the products and calculate its mass that would be produced from
each given reactant. Using H 2 O as the product:
Since H 2 SO 4 produces less H 2 O, it is the limiting reactant and NaOH is in excess.
24. (a) 2 Bi(NO 3 ) 3 + 3 H 2 S ¡ Bi 2 S 3 + 6 HNO 3
50.0 g 6.00 g
Choose one of the products and calculate its mass that would be produced from
each given reactant. Using Bi 2 S 3 as the product:
(b)
110.0 g NaOH2a 1 mol
40.00 g b ¢ 2 mol H 2O
2 mol NaOH ≤ a 18.02 g b = 4.51 g H
mol
2 O
110.0 g H 2 SO 4 2a 1 mol
98.09 g b ¢ 2 mol H 2O
≤ a 18.02 g b = 3.67 g H
1 mol H 2 SO 4 mol
2 O
150.0 g Bi(NO 3 ) 3 )a 1 mol
395.0 g ba 1 mol Bi 2S 3
ba 514.2 g b = 32.5 g Bi
2 mol Bi(NO 3 ) 3 mol
2 S 3
(6.00 g H 2 S)a 1 mol
34.09 g b ¢ 1 mol Bi 2S 3
3 mol H 2 S ≤ a 514.2 g b = 30.2 g Bi
mol
2 S 3
Since H 2 S produces less Bi 2 S 3 , it is the limiting reactant and Bi(NO 3 ) 3 is in excess.
3 Fe + 4 H 2 O ¡ Fe 3 O 4 + 4 H 2
40.0 g 16.0 g
Choose one of the products and calculate its mass that would be produced from
each given reactant. Using H 2 as the product:
(40.0 g Fe)a 1 mol
55.85 g b ¢ 4 mol H 2
3 mol Fe ≤ a 2.016 g b = 1.93 g H
mol
2
116.0 g H 2 O2a 1 mol
18.02 g b ¢ 4 mol H 2
4 mol H 2 O ≤ a 2.016 g b = 1.79 g H
mol
2
Since H 2 O produces less H 2 , it is the limiting reactant and Fe is in excess.
25. Limiting reactant calculations
C 3 H 8 + 5 O 2 ¡ 3 CO 2 + 4 H 2 O
(a) Reaction between 20.0 g C 3 H 8 and 20.0 g O 2
Convert each amount to moles of CO 2
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