\textbf{214 --} Based on the reaction: $3\text{Cu(s)} + 8\text{HNO}_3\text{(aq)} \rightarrow 3\text{Cu(NO}_3)_2\text{(aq)} + 2\text{NO(g)} + 4\text{H}_2\text{O(l)}$, to prepare 14.1 grams of copper(II) nitrate, how many milliliters of 2 molar nitric acid solution are needed? (yield: 80\%)
$$(\text{N} = 14,\ \text{O} = 16,\ \text{Cu} = 64\ \text{: g.mol}^{-1})$$
\hspace{2cm} (1) 125 \hspace{2cm} (2) 100 \hspace{2cm} (3) 50 \hspace{2cm} (4) 25
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