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...
$CuCl 2 +$ type your answer...
$NaNO 3 \to$
type your answer...
$Cu (NO 3) 2 +$
type your answer...
$NaCl$
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?

Answer

Expert–verified

Hide Steps

Answer

Percent\ yield: $\frac{19.3 g NaCl}{34.0024 g NaCl} \cdot 100 = 56.7568 %$

Steps

Step 1 ： ${CuCl}_{2} + 2 {NaNO}_{3} \to Cu ({NO}_{3})_{2} + 2 NaCl$

Step 2 ： $2 NaCl = \frac{25.00 g {CuCl}_{2}}{170.48 g / mol {CuCl}_{2}} \cdot \frac{2 mol NaCl}{1 mol {CuCl}_{2}} \cdot \frac{58.44 g / mol NaCl}{1 mol NaCl} = 34.0024 g NaCl$

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

Step 4 ： $Cu ({NO}_{3})_{2} = \frac{30.00 g {NaNO}_{3}}{85.00 g / mol {NaNO}_{3}} \cdot \frac{1 mol Cu ({NO}_{3})_{2}}{2 mol {NaNO}_{3}} \cdot \frac{187.57 g / mol Cu ({NO}_{3})_{2}}{1 mol Cu ({NO}_{3})_{2}} = 19.9000 g Cu ({NO}_{3})_{2}$

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: $\frac{19.3 g NaCl}{34.0024 g NaCl} \cdot 100 = 56.7568 %$

Answer

Popular Problems