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HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 103<br />
- <strong>Chapter</strong> 9 -<br />
(b)<br />
Aluminum Oxygen<br />
Oxygen is the limiting reactant.<br />
22.<br />
(a)<br />
Nitrogen Oxygen<br />
Oxygen is the limiting reactant.<br />
(b)<br />
Iron Hydrogen Oxygen<br />
Water is the limiting reactant.<br />
23. (a)<br />
KOH + HNO 3 ¡ KNO 3 + H 2 O<br />
16.0 g 12.0 g<br />
Choose one of the products and calculate its mass that would be produced from<br />
each given reactant. Using KNO 3 as the product:<br />
116.0 g KOH2¢ 1 mol<br />
56.10 g ≤ a 1 mol KNO 3<br />
1 mol KOH ≤ a 101.1 g b = 28.8 g KNO<br />
mol<br />
3<br />
112.0 g HNO 3 2a 1 mol<br />
63.02 g b ¢ 1 mol KNO 3<br />
1 mol KOH ≤ a 101.1 g b = 19.3 g KNO<br />
mol<br />
3<br />
Since HNO 3 produces less KNO 3 , it is the limiting reactant and KOH is in excess.<br />
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