10 points
1. When copper (II) chloride reacts with sodium nitrate, copper (II) nitrate and sodium chloride are formed.
a. Write the balanced equation for the reaction given above:
type your answer...
\( \mathrm{CuCl} 2+ \) type your answer...
\( \mathrm{NaNO} 3 \rightarrow \)
type your answer...
\( \mathrm{Cu}(\mathrm{NO} 3) 2+ \)
type your answer...
\( \mathrm{NaCl} \)
b. If 25.00 grams of copper (II) chloride react with 30.00 grams of sodium nitrate, how much sodium chloride can be formed? type your answer...
g
c. What is the limiting reagent for the reaction? type your answer...
d. How many grams of copper (II) nitrate is formed? type your answer...
g
e. How much of the excess reagent is left over in this reaction? type your answer...
g
f. If 19.3 grams of sodium chloride are formed in the reaction described, what is the percent yield of this reaction?

Step 1 ：\( \mathrm{CuCl}_2 + 2 \mathrm{NaNO}_3 \rightarrow \mathrm{Cu}(\mathrm{NO}_3)_2 + 2 \mathrm{NaCl} \)

Step 2 ：\( 2\mathrm{NaCl} = \frac{25.00\,\mathrm{g}\,\mathrm{CuCl}_2}{170.48\,\mathrm{g/mol}\,\mathrm{CuCl}_2} \cdot \frac{2\,\mathrm{mol}\,\mathrm{NaCl}}{1\,\mathrm{mol}\,\mathrm{CuCl}_2} \cdot \frac{58.44\,\mathrm{g/mol}\,\mathrm{NaCl}}{1\,\mathrm{mol}\,\mathrm{NaCl}} = 34.0024\,\mathrm{g}\,\mathrm{NaCl} \)

Step 3 ：Limiting\ reagent: \mathrm{NaNO}_3

Step 4 ：\( \mathrm{Cu}(\mathrm{NO}_3)_2 = \frac{30.00\,\mathrm{g}\,\mathrm{NaNO}_3}{85.00\,\mathrm{g/mol}\,\mathrm{NaNO}_3} \cdot \frac{1\,\mathrm{mol}\,\mathrm{Cu}(\mathrm{NO}_3)_2}{2\,\mathrm{mol}\,\mathrm{NaNO}_3} \cdot \frac{187.57\,\mathrm{g/mol}\,\mathrm{Cu}(\mathrm{NO}_3)_2}{1\,\mathrm{mol}\,\mathrm{Cu}(\mathrm{NO}_3)_2} = 19.9000\,\mathrm{g}\,\mathrm{Cu}(\mathrm{NO}_3)_2 \)

Step 5 ：Excess\ reagent\ left\ over: \mathrm{CuCl}_2 = \frac{30.00\,\mathrm{g}\,\mathrm{NaNO}_3}{85.00\,\mathrm{g/mol}\,\mathrm{NaNO}_3} \cdot \frac{1\,\mathrm{mol}\,\mathrm{CuCl}_2}{2\,\mathrm{mol}\,\mathrm{NaNO}_3} \cdot \frac{170.48\,\mathrm{g/mol}\,\mathrm{CuCl}_2}{1\,\mathrm{mol}\,\mathrm{CuCl}_2} - 25.00\,\mathrm{g}\,\mathrm{CuCl}_2 = -10.8824\,\mathrm{g}\,\mathrm{CuCl}_2 \)

Step 6 ：Percent\ yield: \( \frac{19.3\,\mathrm{g}\,\mathrm{NaCl}}{34.0024\,\mathrm{g}\,\mathrm{NaCl}} \cdot 100 = 56.7568\% \)