Chapter U4, Problem 11RE

(a)

Interpretation Introduction

Interpretation: The type of reaction represented by the given equation is to be determined.

Concept introduction: Acids on reaction with base leads to the formation of salt and water molecule. The amount of product formed in the reaction depends upon the amount of limiting reactant.

Answer to Problem 11RE

The reaction type represented by the given equation is a double displacement reaction.

Explanation of Solution

In the given reaction, ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ reacts with $\text{KOH}$ . In the products, the hydrogen atoms of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ are replaced by potassium atoms and the potassium atoms of $\text{KOH}$ are replaced by a hydrogen atom. Therefore, the given equation represents a double displacement reaction.

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation is to be stated.

Concept introduction: The number of atoms on the left-hand side and right-hand side of the equation is equal in a balanced chemical equation. The amount of product formed in the reaction depends upon the amount of limiting reactant.

Answer to Problem 11RE

The balanced chemical equation is

  ${\text{H}}_{\text{3}}{\text{PO}}_4\left(aq\right)+3\text{KOH}\left(aq\right)\to {\text{K}}_{\text{3}}{\text{PO}}_4\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The given chemical equation is

  ${\text{H}}_{\text{3}}{\text{PO}}_4\left(aq\right)+\text{KOH}\left(aq\right)\to {\text{K}}_{\text{3}}{\text{PO}}_4\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The number of K atom on the right side of the equation is more than the left-hand side of the equation. Whereas the number of H on the left-hand side of the equation is more than the right-hand side of the equation. This equation is balanced by adding a stoichiometric coefficient 3 before $\text{KOH}$ and ${\text{H}}_{\text{2}}\text{O}$ . The balanced chemical equation is

  ${\text{H}}_{\text{3}}{\text{PO}}_4\left(aq\right)+3\text{KOH}\left(aq\right)\to {\text{K}}_{\text{3}}{\text{PO}}_4\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

(c)

Interpretation Introduction

Interpretation: The number of moles of phosphoric acid required to form $5.0\text{mol}$ of potassium phosphate is to be calculated.

Concept introduction: The amount of product formed in the reaction depends upon the amount of limiting reactant. After completion of the chemical reaction, the limiting reagent is fully utilized.

Answer to Problem 11RE

The number of moles of phosphoric acid required to form $5.0\text{mol}$ of potassium phosphate is $5.0\text{mol}$ .

Explanation of Solution

The balanced chemical equation is

  ${\text{H}}_{\text{3}}{\text{PO}}_4\left(aq\right)+3\text{KOH}\left(aq\right)\to {\text{K}}_{\text{3}}{\text{PO}}_4\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The balanced chemical equation shows that the molar ratio between phosphoric acid and potassium phosphate is $1:1$ . Therefore, to form $5.0\text{mol}$ of potassium phosphate, the number of moles of phosphoric acid required is $5.0\text{mol}$ .

(d)

Interpretation Introduction

Interpretation: The maximum amount of potassium phosphate that can be produced from $628.0\text{g}$ of phosphoric acid and $1122.0\text{g}$ of potassium hydroxide is to be calculated.

Concept introduction: The amount of product formed in the reaction depends upon the amount of limiting reactant. After completion of the chemical reaction, the limiting reagent is fully utilized.

Answer to Problem 11RE

The maximum amount of potassium phosphate that can be produced from $628.0\text{g}$ of phosphoric acid and $1122.0\text{g}$ of potassium hydroxide is $1360.65\text{g}$ .

Explanation of Solution

The number of moles of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is calculated as,

  $n_{{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}=\frac{m_{{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}}{M_{{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}}$

Where,

• $m_{{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}$ is the given mass of phosphoric acid.
• $M_{{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}$ is the molar mass of phosphoric acid.

The molar mass of phosphoric acid is $97.994\text{g/mol}$ .

Substitute the given mass and molar mass of phosphoric acid in the above formula.

  $\begin{array}{c}{n}_{{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}=\frac{628.0\text{g}}{97.994\text{g/mol}}\\ =6.41\text{mol}\end{array}$

The number of moles of $\text{KOH}$ is calculated as,

  $n_{\text{KOH}}=\frac{m_{\text{KOH}}}{M_{\text{KOH}}}$

Where,

• $m_{\text{KOH}}$ is the given mass of potassium hydroxide.
• $M_{\text{KOH}}$ is the molar mass of potassium hydroxide.

The molar mass of potassium hydroxide is $56.1056\text{g/mol}$ .

Substitute the given mass and molar mass of potassium hydroxide in the above formula.

  $\begin{array}{c}{n}_{\text{KOH}}=\frac{1122.0\text{g}}{56.1056\text{g/mol}}\\ =20\text{mol}\end{array}$

The molar ratio of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ and $\text{KOH}$ is $1:3$ . Therefore, phosphoric acid is the limiting reactant.

The balanced chemical equation shows that the molar ratio between phosphoric acid and potassium phosphate is $1:1$ . Therefore, the number of moles of potassium phosphate produced by $6.41\text{mol}$ of phosphoric acid is $6.41\text{mol}$ .

The amount of potassium phosphate is calculated as,

  $m_{{\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}}=n_{{\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}}\times M_{{\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}}$

Where,

• $n_{{\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}}$ is the number of moles of potassium phosphate.
• $M_{{\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}}$ is the molar mass of potassium phosphate $\left(212.27\text{g/mol}\right)$ .

Substitute the number of moles and molar mass of potassium phosphate in the above formula.

  $\begin{array}{c}{m}_{{\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}}=6.41\text{mol}\times 212.27\text{g/mol}\\ =1360.65\text{g}\end{array}$

(e)

Interpretation Introduction

Interpretation: The limiting reactant of part d and its amount of excess reactant in moles and grams is to be determined.

Concept introduction: The amount of product formed in the reaction depends upon the amount of limiting reactant. After completion of the chemical reaction, the limiting reagent is fully utilized.

Answer to Problem 11RE

Phosphoric acid is the limiting reactant. The amount of excess reactant in moles and grams is $0.77\text{mol}$ and $43.20\text{g}$ , respectively.

Explanation of Solution

The number of moles of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ and $\text{KOH}$ is $6.41\text{mol}$ and $20\text{mol}$ . The molar ratio of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ and $\text{KOH}$ is $1:3$ . Therefore, phosphoric acid is the limiting reactant.

The number of moles of $\text{KOH}$ that completely reacts with $6.41\text{mol}$ of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $19.23\text{mol}$ . The number of moles of $\text{KOH}$ left is $0.77\text{mol}$ .

The amount $\text{KOH}$ in grams is calculated as,

The number of moles of $\text{KOH}$ is calculated as,

  $m_{\text{KOH}}=n_{\text{KOH}}\times M_{\text{KOH}}$

Where,

• $n_{\text{KOH}}$ is the number of moles of potassium hydroxide.
• $M_{\text{KOH}}$ is the molar mass of potassium hydroxide.

The molar mass of potassium hydroxide is $56.1056\text{g/mol}$ .

Substitute the number of moles and molar mass of potassium hydroxide in the above formula.

  $\begin{array}{c}{m}_{\text{KOH}}=0.77\text{mol}\times 56.1056\text{g/mol}\\ =43.20\text{g}\end{array}$