Chapter 12, Problem 72A

a)

Interpretation Introduction

Interpretation: The balanced equation for the reaction is to be written.

Concept Introduction: Reactants, products, and an arrow indicating the reaction's direction make up a chemical equation. A balanced chemical equation is one in which the number of atoms in each of the molecules is the same on both sides of the equation.

Answer to Problem 72A

The balanced chemical equation is given as

  $\begin{array}{l}{\text{2Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}{\text{+6SiO}}_{\text{2} }\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{+6CaSiO}}_{\text{3}}\hfill \\ {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\text{+10C }\to {\text{P}}_{\text{4}}\text{+10CO}\hfill \end{array}$

Explanation of Solution

The given chemical reaction is as follows:

  $\begin{array}{l}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}{\text{+SiO}}_{\text{2} }\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{+CaSiO}}_{\text{3}}\hfill \\ {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\text{+C }\to {\text{P}}_{\text{4}}\text{+CO}\hfill \end{array}$

The balanced chemical equation is given as

  $\begin{array}{l}{\text{2Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}{\text{+6SiO}}_{\text{2} }\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{+6CaSiO}}_{\text{3}}\hfill \\ {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\text{+10C }\to {\text{P}}_{\text{4}}\text{+10CO}\hfill \end{array}$

b)

Interpretation Introduction

Interpretation: The limiting reagent is to be identified.

Concept Introduction: The reactant that limits how much product can be produced by a reaction is known as the limiting reagent. Only until the limiting reagent is consumed does the reaction take place.

Answer to Problem 72A

The limiting reagent is ${\text{SiO}}_{\text{2} }$ .

Explanation of Solution

Mass of calcium phosphate is $5.5\times {10}^5\text{g}$ .

Mass of sand is $2.3\times {10}^5\text{g}$ .

Molar mass of calcium phosphate is $310.3\text{g}$ .

Molar mass of ${\text{SiO}}_{\text{2}}$ is $60.1\text{g}$ .

The balanced chemical equation is given as

  $\begin{array}{l}{\text{2Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}{\text{+6SiO}}_{\text{2} }\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{+6CaSiO}}_{\text{3}}\hfill \\ {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\text{+10C }\to {\text{P}}_{\text{4}}\text{+10CO}\hfill \end{array}$

  $2\text{mol}$ of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ reacts with $1\text{mol}$ of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ .

  $6\text{mol}$ of ${\text{SiO}}_{\text{2} }$ reacts with $1\text{mol}$ of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ .

The moles of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ reacted with respect to calcium phosphate is given as follows:

  $5.5\times {10}^5\cancel{\text{g}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}}\times \frac{1\cancel{\text{mol}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}}}{310.3\cancel{\text{g}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}}}\times \frac{1\text{mol}{\text{P}}_4{\text{O}}_{10}}{2\cancel{\text{mol}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}}}=8.9\times {10}^2\text{mol}{\text{P}}_4{\text{O}}_{10}$

The moles of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ reacted with respect to ${\text{SiO}}_{\text{2} }$ is given as follows:

  $2.3\times {10}^5\cancel{\text{g}{\text{SiO}}_{\text{2} }}\times \frac{1\cancel{\text{mol}{\text{SiO}}_{\text{2} }}}{60.1\cancel{\text{g}{\text{SiO}}_{\text{2} }}}\times \frac{1\text{mol}{\text{P}}_4{\text{O}}_{10}}{6\cancel{\text{mol}{\text{SiO}}_{\text{2} }}}=6.4\times {10}^2\text{mol}{\text{P}}_4{\text{O}}_{10}$

Since moles of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ consumed for ${\text{SiO}}_{\text{2} }$ is less, ${\text{SiO}}_{\text{2} }$ is the limiting reagent.

c)

Interpretation Introduction

Interpretation: The grams of phosphorus produced is to be calculated.

Concept Introduction: Reactants, products, and an arrow indicating the reaction's direction make up a chemical equation. A balanced chemical equation is one in which the number of atoms in each of the molecules is the same on both sides of the equation.

Answer to Problem 72A

The grams of phosphorus produced is $7.9\times {10}^4\text{g}$ .

Explanation of Solution

Mass of sand is $2.3\times {10}^5\text{g}$ .

Molar mass of ${\text{SiO}}_{\text{2}}$ is $60.1\text{g}$ .

The balanced chemical equation is given as

  $\begin{array}{l}{\text{2Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}{\text{+6SiO}}_{\text{2} }\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{+6CaSiO}}_{\text{3}}\hfill \\ {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\text{+10C }\to {\text{P}}_{\text{4}}\text{+10CO}\hfill \end{array}$

  $6\text{mol}$ of ${\text{SiO}}_{\text{2} }$ reacts with $1\text{mol}$ of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ .

The moles of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ reacted with respect to ${\text{SiO}}_{\text{2} }$ is given as follows:

  $2.3\times {10}^5\cancel{\text{g}{\text{SiO}}_{\text{2} }}\times \frac{1\cancel{\text{mol}{\text{SiO}}_{\text{2} }}}{60.1\cancel{\text{g}{\text{SiO}}_{\text{2} }}}\times \frac{1\text{mol}{\text{P}}_4{\text{O}}_{10}}{6\cancel{\text{mol}{\text{SiO}}_{\text{2} }}}=6.4\times {10}^2\text{mol}{\text{P}}_4{\text{O}}_{10}$

  $1\text{mol}$ of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ gives $1\text{mol}$ of ${\text{P}}_{\text{4}}$ .

The grams of ${\text{P}}_{\text{4}}$

produced is given as follows:

  $6.4\times {10}^2\cancel{\text{mol}{\text{P}}_4{\text{O}}_{10}}\times \frac{1\cancel{\text{mol}{\text{P}}_4}}{1\cancel{\text{mol}{\text{P}}_4{\text{O}}_{10}}}\times \frac{124.0\text{g}\text{}{\text{P}}_4}{1\cancel{\text{mol}{\text{P}}_4}}=7.9\times {10}^4\text{g}{\text{P}}_4$

d)

Interpretation Introduction

Interpretation: The grams of carbon consumed is to be calculated.

Concept Introduction: Reactants, products, and an arrow indicating the reaction's direction make up a chemical equation. A balanced chemical equation is one in which the number of atoms in each of the molecules is the same on both sides of the equation.

Answer to Problem 72A

The grams of carbon consumed is $7.7\times {10}^4\text{g}$ .

Explanation of Solution

Mass of sand is $2.3\times {10}^5\text{g}$ .

Molar mass of ${\text{SiO}}_{\text{2}}$ is $60.1\text{g}$ .

The balanced chemical equation is given as

  $\begin{array}{l}{\text{2Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}{\text{+6SiO}}_{\text{2} }\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{+6CaSiO}}_{\text{3}}\hfill \\ {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\text{+10C }\to {\text{P}}_{\text{4}}\text{+10CO}\hfill \end{array}$

  $6\text{mol}$ of ${\text{SiO}}_{\text{2} }$ reacts with $1\text{mol}$ of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ .

The moles of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ reacted with respect to ${\text{SiO}}_{\text{2} }$ is given as follows:

  $2.3\times {10}^5\cancel{\text{g}{\text{SiO}}_{\text{2} }}\times \frac{1\cancel{\text{mol}{\text{SiO}}_{\text{2} }}}{60.1\cancel{\text{g}{\text{SiO}}_{\text{2} }}}\times \frac{1\text{mol}{\text{P}}_4{\text{O}}_{10}}{6\cancel{\text{mol}{\text{SiO}}_{\text{2} }}}=6.4\times {10}^2\text{mol}{\text{P}}_4{\text{O}}_{10}$

  $1\text{mol}$ of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ reacts with $10\text{mol}$ of $\text{C}$ .

The grams of carbon consumed is given as follows:

  $6.4\times {10}^2\cancel{\text{mol}{\text{P}}_4{\text{O}}_{10}}\times \frac{10\cancel{\text{mol}\text{C}}}{1\cancel{\text{mol}{\text{P}}_4{\text{O}}_{10}}}\times \frac{12.0\text{g}\text{}\text{C}}{1\cancel{\text{mol}\text{C}}}=7.7\times {10}^4\text{g}\text{C}$