Chapter 12, Problem 56A

a.

Interpretation Introduction

Interpretation: To calculate the amount of phosphoric acid 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. The coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

Answer to Problem 56A

The amount of phosphoric acid is $2.35\text{ g}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  $3{\text{ CaCO}}_3\left(s\right)+2{\text{H}}_3{\text{PO}}_4\left(aq\right)\to {\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(aq\right)+3{\text{CO}}_{\text{2}}\left(g\right)+3{\text{ H}}_2\text{O}\left(l\right)$

When three moles of calcium carbonate react with two moles of phosphoric acid it yields one mole of calcium phosphate, three moles of carbon dioxide, and three moles of water.

The amount of calcium phosphate is $3.74\text{ g}$ .

Therefore, the moles of calcium phosphate can be calculated as follows:

  $3.74\cancel{{\text{g Ca}}_3{\left({\text{PO}}_4\right)}_2}\times \frac{1{\text{ mol Ca}}_3{\left({\text{PO}}_4\right)}_2}{\text{310}\text{.174 }\cancel{{\text{g Ca}}_3{\left({\text{PO}}_4\right)}_2}}=0.012{\text{mol Ca}}_3{\left({\text{PO}}_4\right)}_2$

As per the stoichiometry ratio, the number of moles of phosphoric acid is as follows:

  $2\times 0.012\text{mol}=0.024{\text{mol H}}_3{\text{PO}}_4$

Calculation for the amount of phosphoric acid in grams is as follows:

  $0.024\cancel{{\text{mol H}}_3{\text{PO}}_4}\times \frac{97.994{\text{ g H}}_3{\text{PO}}_4}{\text{1 }\cancel{{\text{mol H}}_3{\text{PO}}_4}}=2.35{\text{ g H}}_3{\text{PO}}_4$

b.

Interpretation Introduction

Interpretation: To calculate the amount of carbon dioxide.

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. The coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

Answer to Problem 56A

The amount of carbon dioxide is $1.88\text{ g}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  $3{\text{ CaCO}}_3\left(s\right)+2{\text{H}}_3{\text{PO}}_4\left(aq\right)\to {\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(aq\right)+3{\text{CO}}_{\text{2}}\left(g\right)+3{\text{ H}}_2\text{O}\left(l\right)$

When three moles of calcium carbonate react with two moles of phosphoric acid, it yields one mole of calcium phosphate, three moles of carbon dioxide, and three moles of water.

The amount of water is $0.773\text{ g}$ .

The amount of carbon dioxide can be calculated as follows:

  $0.773\cancel{{\text{g H}}_2\text{O}}\times \frac{1\cancel{{\text{mol H}}_2\text{O}}}{18\cancel{{\text{g H}}_2\text{O}}}\times \frac{3\cancel{{\text{mol CO}}_2}}{3\cancel{{\text{mol H}}_2\text{O}}}\times \frac{{\text{44 g CO}}_2}{1\cancel{{\text{mol CO}}_2}}=1.88{\text{ g CO}}_2$