Chapter U4, Problem 12RE

(a)

Interpretation Introduction

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

Concept introduction:The reaction in which displacement of ions take place between two reactant molecules is known as double displacement reaction.

Answer to Problem 12RE

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

Explanation of Solution

The given reaction is shown below:

  ${\text{K}}_{\text{3}}{\text{PO}}_4\left(aq\right)+{\text{H}}_2{\text{SO}}_4\left(aq\right)\to {\text{H}}_{\text{3}}{\text{PO}}_4\left(aq\right)+{\text{K}}_{\text{2}}{\text{SO}}_4\left(aq\right)$

In the given reaction, ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ reactswith ${\text{H}}_2{\text{SO}}_4$ . The potassium atoms of ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ are replaced by hydrogen atoms and hydrogen atom of ${\text{H}}_2{\text{SO}}_4$ is replaced by potassium atom. Therefore, the given equation represents 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 balanced chemical equation.

Answer to Problem 12RE

The balanced chemical equation is

  $2{\text{K}}_{\text{3}}{\text{PO}}_4\left(aq\right)+3{\text{H}}_2{\text{SO}}_4\left(aq\right)\to 2{\text{H}}_{\text{3}}{\text{PO}}_4\left(aq\right)+3{\text{K}}_{\text{2}}{\text{SO}}_4\left(aq\right)$

Explanation of Solution

The given chemical equation is

  ${\text{K}}_{\text{3}}{\text{PO}}_4\left(aq\right)+{\text{H}}_2{\text{SO}}_4\left(aq\right)\to {\text{H}}_{\text{3}}{\text{PO}}_4\left(aq\right)+{\text{K}}_{\text{2}}{\text{SO}}_4\left(aq\right)$

The number of hydrogen atom and potassium atom on both sides of the equation are not equal.The balanced chemical equation is

  $2{\text{K}}_{\text{3}}{\text{PO}}_4\left(aq\right)+3{\text{H}}_2{\text{SO}}_4\left(aq\right)\to 2{\text{H}}_{\text{3}}{\text{PO}}_4\left(aq\right)+3{\text{K}}_{\text{2}}{\text{SO}}_4\left(aq\right)$

(c)

Interpretation Introduction

Interpretation: The number of moles of potassium phosphate required to form $6\text{mol}$ of phosphoric acid is to be calculated.

Concept introduction: The amount of product formed in the reaction depends upon the amount of limiting reactant.

Answer to Problem 12RE

The number of moles of potassium phosphate required to form $6\text{mol}$ of phosphoric acidis $6\text{mol}$ .

Explanation of Solution

The balanced chemical equation is

  $2{\text{K}}_{\text{3}}{\text{PO}}_4\left(aq\right)+3{\text{H}}_2{\text{SO}}_4\left(aq\right)\to 2{\text{H}}_{\text{3}}{\text{PO}}_4\left(aq\right)+3{\text{K}}_{\text{2}}{\text{SO}}_4\left(aq\right)$

The balanced chemical equation shows that the molar ratio between potassium phosphate and phosphoric acid is $2:2$ or $1:1$ . Therefore, in order to form $6\text{mol}$ of phosphoric acid, the number of moles of potassium phosphate required is $6\text{mol}$ .

(d)

Interpretation Introduction

Interpretation: The number of moles of sulfuric acid required to form $6\text{mol}$ of phosphoric acid is to be calculated.

Concept introduction: The amount of product formed in the reaction depends upon the amount of limiting reactant.

Answer to Problem 12RE

The number of moles of sulfuric acid required to form $6\text{mol}$ of phosphoric acidis $9\text{mol}$ .

Explanation of Solution

The balanced chemical equation is

  $2{\text{K}}_{\text{3}}{\text{PO}}_4\left(aq\right)+3{\text{H}}_2{\text{SO}}_4\left(aq\right)\to 2{\text{H}}_{\text{3}}{\text{PO}}_4\left(aq\right)+3{\text{K}}_{\text{2}}{\text{SO}}_4\left(aq\right)$

The balanced chemical equation shows that the molar ratio between sulfuric acid and phosphoric acid is $3:2$ . Therefore, in order to form $6\text{mol}$ of phosphoric acid, the number of moles of sulfuric acid required is $9\text{mol}$ .

(e)

Interpretation Introduction

Interpretation: The maximum number of moles of phosphoric acid that can be produced from $750.0\text{g}$ of potassium phosphate and $0.75\text{L}$ of $18.4\text{M}$ sulfuric acid is to be calculated.

Concept introduction: The amount of product formed in the reaction depends upon the amount of limiting reactant.

Answer to Problem 12RE

The maximum number of moles of phosphoric acid that can be produced from $750.0\text{g}$ of potassium phosphate and $0.75\text{L}$ of $18.4\text{M}$ sulfuric acid is $3.53\text{mol}$ .

Explanation of Solution

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

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

Where,

• $m_{{\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}}$ is the given mass of potassium phosphate.
• $M_{{\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}}$ is the molar mass of potassium phosphate.

The molar mass of potassium phosphateis $212.27\text{g/mol}$ .

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

  $\begin{array}{c}{n}_{{\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}}=\frac{750.0\text{g}}{212.27\text{g/mol}}\\ =3.53\text{mol}\end{array}$

The number of moles of sulfuric acid is calculated as,

  $n_{{\text{H}}_2{\text{SO}}_4}=M\times V$

Where,

• M is the molarity of sulfuric acid.
• V is the given volume of sulfuric acid.

Substitute the given molarity and volume ofsulfuric acid in the above formula.

  $\begin{array}{c}{n}_{{\text{H}}_2{\text{SO}}_4}=18.4\text{M}\times 0.75\text{L}\\ =13.8\text{mol}\end{array}$

The molar ratio of ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $2:3$ .The number of moles of sulfuric acid used by $3.53\text{mol}$ of potassium phosphate is

  $\begin{array}{c}n=\frac{3}{2}\times 3.53\text{mol}\\ =5.295\text{mol}\end{array}$

Therefore, potassium phosphate is the limiting reactant.

The molar ratio between potassium phosphate and phosphoric acid is $1:1$ .

Therefore, the maximum number of moles of phosphoric acid that can be produced is $3.53\text{mol}$ .

(f)

Interpretation Introduction

Interpretation: The limiting reactant of part e 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.

Answer to Problem 12RE

Potassium phosphate is the limiting reactant. The amount of excess reactant in moles and grams is $8.505\text{mol}$ and $834.16\text{g}$ , respectively.

Explanation of Solution

The number of moles of ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ used in the reaction is $3.53\text{mol}$ and $5.295\text{mol}$ , respectively. The molar ratio of ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $2:3$ . Therefore, potassium phosphate is the limiting reactant.

The number of moles of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ that completely reacts with $3.53\text{mol}$ of ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $5.295\text{mol}$ . The number of moles of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ left is $8.505\text{mol}$ .

The amount ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ in grams is calculated as,

  $m_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}=n_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\times M_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}$

Where,

• $n_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}$ is the number of moles of sulfuric acid.
• $M_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}$ is the molar mass of sulfuric acid.

The molar mass of sulfuric acidis $98.079\text{g/mol}$ .

Substitute the number of moles and molar mass of sulfuric acid in the above formula.

  $\begin{array}{c}{m}_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}=8.505\text{mol}\times 98.079\text{g/mol}\\ =834.16\text{g}\end{array}$