Question

reacts with hcl.

1. zinc reacts with iodine in a synthesis reaction: $ce{zn + i_{2} -> zni_{2}}$

a. determine the theoretical yield if 1.912 mol of zinc is used.
b. determine the percent yield if 515.6 g of product is recovered.

1. challenge when copper wire is placed into a silver nitrate solution ($ce{agno_{3}}$), silver crystals and copper(ii) nitrate ($ce{cu(no_{3})_{2}}$) solution form.

a. write the balanced chemical equation for the reaction.
b. if a 20.0-g sample of copper is used, determine the theoretical yield of silver.
c. if 60.0 g of silver is recovered from the reaction, determine the percent yield of the reaction.

Explanation:

Problem 29

Step1: Mole ratio of Zn to ZnI₂

From the balanced equation $\text{Zn} + \text{I}_2
ightarrow \text{ZnI}_2$, mole ratio $\text{Zn}:\text{ZnI}_2 = 1:1$.
Moles of $\text{ZnI}_2$ = moles of $\text{Zn} = 1.912\ \text{mol}$
Molar mass of $\text{ZnI}_2$: $65.38 + 2\times126.90 = 319.18\ \text{g/mol}$
Theoretical yield mass: $1.912\ \text{mol} \times 319.18\ \text{g/mol}$

Step2: Calculate theoretical yield

$1.912 \times 319.18 = 610.27\ \text{g}$

Step3: Calculate percent yield

Percent yield formula: $\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$
Substitute values: $\frac{515.6\ \text{g}}{610.27\ \text{g}} \times 100\%$

Step4: Compute final percent yield

$\frac{515.6}{610.27} \times 100\% \approx 84.49\%$

Step1: Balance the chemical equation

Unbalanced: $\text{Cu} + \text{AgNO}_3
ightarrow \text{Ag} + \text{Cu(NO}_3\text{)}_2$
Balance $\text{NO}_3^-$ and $\text{Ag}$: $\text{Cu} + 2\text{AgNO}_3
ightarrow 2\text{Ag} + \text{Cu(NO}_3\text{)}_2$

Step2: Calculate moles of Cu

Molar mass of $\text{Cu} = 63.55\ \text{g/mol}$
Moles of $\text{Cu} = \frac{20.0\ \text{g}}{63.55\ \text{g/mol}}$

Step3: Find moles of Ag (mole ratio 1:2)

Moles of $\text{Ag} = 2 \times \frac{20.0}{63.55} = \frac{40.0}{63.55} \approx 0.6294\ \text{mol}$
Molar mass of $\text{Ag} = 107.87\ \text{g/mol}$
Theoretical yield of $\text{Ag}$: $0.6294\ \text{mol} \times 107.87\ \text{g/mol}$

Step4: Compute theoretical Ag mass

$0.6294 \times 107.87 \approx 67.9\ \text{g}$

Step5: Calculate percent yield of Ag

Percent yield formula: $\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$
Substitute values: $\frac{60.0\ \text{g}}{67.9\ \text{g}} \times 100\%$

Step6: Compute final percent yield

$\frac{60.0}{67.9} \times 100\% \approx 88.4\%$

Answer:

a. $610.3\ \text{g}$ (rounded to 4 significant figures)
b. $84.5\%$ (rounded to 3 significant figures)

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Problem 30