Chapter 12, Problem 57A

a.

Interpretation Introduction

Interpretation: To calculate the number of atoms of zinc.

Concept Introduction: A mole is $6.02\times {10}^{23}$ atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction.

Answer to Problem 57A

The number of atoms of zinc are $5.70\times {10}^{21}$ atoms.

Explanation of Solution

The given chemical reaction is depicted as follows:

  $4\text{ Zn}\left(s\right)+10{\text{HNO}}_3\left(aq\right)\to 4\text{Zn}{\left({\text{NO}}_3\right)}_{\text{2}}\left(aq\right)+{\text{NH}}_4{\text{NO}}_3\left(aq\right)+3{\text{ H}}_2\text{O}\left(l\right)$

When four moles of zinc react with ten moles of nitric acid, it yields four moles of zinc nitrate along with ammonium nitrate and three moles of water.

The amount of nitric acid is $1.49\text{ g}$ .

Calculation for the number of atoms of zinc is as follows:

  $1.49\cancel{{\text{g HNO}}_3}\times \frac{1\cancel{{\text{mol HNO}}_3}}{63.0\cancel{{\text{g HNO}}_3}}\times \frac{4\cancel{\text{mol Zn}}}{10\cancel{{\text{mol HNO}}_3}}\times \frac{6.022\times {10}^{23}\text{ atoms Zn}}{1\cancel{\text{mol Zn}}}=5.70\times {10}^{21}\text{ atoms Zn}$

b.

Interpretation Introduction

Interpretation: To calculate the amount of zinc in grams.

Concept Introduction: A mole is $6.02\times {10}^{23}$ atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction.

Answer to Problem 57A

The amount of zinc in grams is $95.2\text{ g}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  $4\text{ Zn}\left(s\right)+10{\text{HNO}}_3\left(aq\right)\to 4\text{Zn}{\left({\text{NO}}_3\right)}_{\text{2}}\left(aq\right)+{\text{NH}}_4{\text{NO}}_3\left(aq\right)+3{\text{ H}}_2\text{O}\left(l\right)$

When four moles of zinc react with ten moles of nitric acid, it yields four moles of zinc nitrate along with ammonium nitrate and three moles of water.

The amount of ammonium nitrate is $29.1\text{ g}$ .

The amount of zinc can be calculated as follows:

  $29.1\cancel{{\text{g NH}}_4{\text{NO}}_3}\times \frac{1\cancel{{\text{mol NH}}_4{\text{NO}}_3}}{80.0\cancel{{\text{g NH}}_4{\text{NO}}_3}}\times \frac{4\cancel{\text{mol Zn}}}{1\cancel{{\text{mol NH}}_4{\text{NO}}_3}}\times \frac{\text{65}\text{.4 g Zn}}{1\cancel{\text{mol Zn}}}=95.2\text{ g Zn}$