Chapter 12, Problem 62A

Interpretation Introduction

Interpretation : The kilograms of ${\text{CaSO}}_{\text{4}}$ formed is to be calculated.

Concept Introduction : The coefficients of a balanced chemical equation that are transformed into moles are used to calculate the mole ratio, which is a conversion factor.

When converting between a certain number of moles of one reactant or product and moles of another reactant or product, mole ratios are utilized.

The formula to calculate the number of moles is given as

  $\text{n}\text{=}\frac{\text{given}\text{mass}}{\text{molar mass}}$   ...... (1)

The formula to calculate the percentage yield is given as

  $\text{Percent}\text{yield =}\frac{\text{actual}\text{yield}}{\text{theoretical}\text{yield}}\times 100\%$   ...... (2)

Answer to Problem 62A

The kilograms of ${\text{CaSO}}_{\text{4}}$ formed is $10.778\text{kg}$ .

Explanation of Solution

Given information:

The yield is $96.8\%$ .

The amount of ${\text{SO}}_{\text{2}}$ reacted is $\text{5}\text{.24 kg}$ .

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

The given chemical equation is as follows:

  $2{\text{CaCO}}_{\text{3}}\left(s\right)+2{\text{SO}}_{\text{2}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right) \to 2{\text{CaSO}}_{\text{4}}\left(s\right)+2{\text{CO}}_{\text{2}}\left(g\right)$

Molar mass of S is $\text{32}\text{.07}\text{g/mol}$ .

Molar mass of O is $\text{16}\text{.00}\text{g/mol}$ .

Molar mass of $\text{Ca}$ is $\text{40}\text{.08}\text{g/mol}$ .

Mass of ${\text{SO}}_{\text{2}}$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{SO}}_{\text{2}}=\text{M}\left(\text{S}\right)\text{+2M}\left(\text{O}\right)\\ =\left(32.07+\left(2\times 16.00\right)\right)\text{g/mol}\\ \text{= 64}\text{.02}\text{g/mol}\end{array}$

Mass of ${\text{CaSO}}_{\text{4}}$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{CaSO}}_{\text{4}}=\text{M}\left(\text{Ca}\right)\text{+M}\left(\text{S}\right)+4\text{M}\left(\text{O}\right)\\ =\left(40.08+32.07+\left(4\times 16.00\right)\right)\text{g/mol}\\ \text{= 136}\text{.15}\text{g/mol}\end{array}$

To calculate the moles of ${\text{SO}}_{\text{2}}$ , substitute the values in equation (1).

  $\begin{array}{c}{\text{n}}_{{\text{SO}}_2}\text{=}\frac{{\text{m}}_{{\text{SO}}_2}}{\text{M}\left({\text{SO}}_2\right)}\\ =\frac{5.24\times {10}^3\cancel{\text{g}}}{64.02\cancel{\text{g}}\text{/mol}}\\ =81.7855\text{mol}\end{array}$

The moles of ${\text{CaSO}}_4$ is equal to the moles of ${\text{SO}}_{\text{2}}$ .

So, the moles of ${\text{CaSO}}_4$ is $81.7855\text{mol}$ .

To calculate the mass of ${\text{CaSO}}_4$ , substitute the values in equation (1).

  $\begin{array}{c}{\text{m}}_{{\text{CaSO}}_4}={\text{n}}_{{\text{CaSO}}_4}\times \text{M}\left({\text{CaSO}}_4\right)\\ =81.7855\cancel{\text{mol}}\times 136.15\text{g/}\cancel{\text{mol}}\\ =1.1135\times {10}^4\text{g}\end{array}$

The theoretical mass of ${\text{CaSO}}_4$ is $1.1135\times {10}^4\text{g}$ .

To calculate the actual mass of ${\text{CaSO}}_4$ , substitute the values in equation (2).

  $\begin{array}{c}\text{Percent}\text{yield}\text{=}\frac{\text{actual}\text{yield}}{\text{theoretical}\text{yield}}\times 100\%\\ \text{actual yield}\text{=}\frac{\text{percent}\text{yield}\times \text{theoretical yield}}{100\%}\\ =\frac{96.8}{100}\times 1.1135\times {10}^4\text{g}\left(\because \text{1}\text{kg}\text{=}{\text{10}}^3\text{g}\right)\\ =10.778\text{kg}\end{array}$