Chapter 11.4, Problem 29PP

(a)

Interpretation Introduction

Interpretation:

The theoretical yield if 1.912 moles of Zn is used in the given reaction needs to be determined.

${\text{Zn + I}}_{\text{2}}^{}\stackrel{}{\to }{\text{ ZnI}}_{\text{2}}$

Concept introduction:

The actual yield of the product is the amount of product, whichform at the end of reaction experimentally whereas theoretical yield represents the amount of product calculated by expecting that 100 % of reactant change to product. The percent yield can be shown as:

$\text{Percent yield = }\frac{\text{Actual yield}}{\text{Theoretical yield}}\text{ x 100}$

Answer to Problem 29PP

Theoretical yield = $610gZn{I}_2$

Explanation of Solution

Moles of Zn = 1.912 moles

Molar mass of ${\text{ZnI}}_{\text{2}}$ = 319 g/mol

${\text{Zn + I}}_{\text{2}}^{}\stackrel{}{\to }{\text{ ZnI}}_{\text{2}}$

Calculate mass of ${\text{ZnI}}_{\text{2}}$ :

$1.912molesZn\times \frac{1moleZn{I}_2}{1moleZn}\times \frac{319gZn{I}_2}{1moleZn{I}_2}=610gZn{I}_2$

(b)

Interpretation Introduction

Interpretation:

The percent yield if theoretical yield is 610 g for the reaction of 1.912 moles of Zn is used in the given reaction needs to be determined.

${\text{Zn + I}}_{\text{2}}^{}\stackrel{}{\to }{\text{ ZnI}}_{\text{2}}$

Concept introduction:

The actual yield of the product is the amount of product, which form at the end of reaction experimentally, whereas theoretical yield represents the amount of product calculated by expecting that 100 % of reactant change to product. The percent yield can be shown as:

$\text{Percent yield = }\frac{\text{Actual yield}}{\text{Theoretical yield}}\text{ x 100}$

Answer to Problem 29PP

Percent yield = $\text{84}\text{.5 \%}$

Explanation of Solution

Moles of Zn = 1.912 moles

Theoretical yield = $610gZn{I}_2$

Actual yield = 515.6 g

Calculate mass of percent yield:

$\text{Percent yield = }\frac{\text{515}\text{.6 g}}{\text{610 g}}\text{ x 100 = 84}\text{.5 \%}$