## Chemistry 10th Edition

(c) $4Zn + 10HNO_3 \longrightarrow 4Zn(NO_3)_2+NH_4NO_3+3H_2O$
1. Find how many mol of $HNO_3$ is necessary for each case: a. $$6.5 \space mol \space H_2O \times \frac{ 8 \space mol \ HNO_3 }{ 4 \space mol \space H_2O } = 13 \space mol \space HNO_3$$ b. $$6.5 \space mol \space H_2O \times \frac{ 6 \space mol \ HNO_3 }{ 3 \space mol \space H_2O } = 13 \space mol \space HNO_3$$ c. $$6.5 \space mol \space H_2O \times \frac{ 10 \space mol \ HNO_3 }{ 3 \space mol \space H_2O } = 22 \space mol \space HNO_3$$ - As we can see, reaction (c) uses the most nitric acid to produce 6.5 mol of water.