Chapter 10, Problem 8PE

Interpretation Introduction

Interpretation:

The number of kilograms of sulfur dioxide that will result from burning $5.00\times {10}^2$ kilograms of carbon disulfide is to be calculated.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in a reaction which determines the moles of the reactants and products in the reaction.

Answer to Problem 8PE

The number of kilograms of sulfur dioxide that will result from burning $5.00\times {10}^2$ kilograms of carbon disulfide is $782\text{kg}$.

Explanation of Solution

The balanced equation for the reaction is given below.

${\text{CS}}_2\left(\text{l}\right){\text{+3O}}_2\left(\text{g}\right)\to 2{\text{SO}}_2\left(\text{g}\right)+{\text{CO}}_2\left(\text{g}\right)$

Therefore, $1$ kilomole of ${\text{CS}}_2$ produces $\text{2}$ kilomoles of ${\text{SO}}_2$.

Therefore kilomole to kilomole ratio is given below.

$\text{2}\text{kmol}{\text{SO}}_2=1\text{kmol}{\text{CS}}_2$

Therefore, two conversion factors from the kilomole-to-kilomole ratio are given below.

$\frac{\text{2}\text{kmol}{\text{SO}}_2}{1\text{kmol}{\text{CS}}_2}\text{and}\frac{1\text{kmol}{\text{CS}}_2}{\text{2}\text{kmol}{\text{SO}}_2}$

The conversion factor to obtain kilomoles of ${\text{SO}}_2$ from ${\text{CS}}_2$ is given below.

$\frac{\text{2}\text{kmol}{\text{SO}}_2}{1\text{kmol}{\text{CS}}_2}$

The molar mass of carbon is $12.01\text{kg}{\text{mol}}^{-1}$.

The molar mass of sulfur is $32.06\text{kg}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{CS}}_2$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 12.01\text{kg}{\text{mol}}^{-1}+\left(2\times 32.06\text{kg}{\text{mol}}^{-1}\right)\\ & =& 12.01\text{kg}{\text{mol}}^{-1}+64.12\text{kg}{\text{mol}}^{-1}\\ & =& 76.13\text{kg}{\text{mol}}^{-1}\end{array}$

Therefore, the conversion factor to obtain kilomoles of ${\text{CS}}_2$ from kilograms of ${\text{CS}}_2$ is given below.

$\frac{\text{1}\text{kmol}{\text{CS}}_2}{\text{76}\text{.13}\text{kg}{\text{CS}}_2}$

The molar mass of oxygen is $16.00\text{kg}{\text{mol}}^{-1}$.

The molar mass of sulfur is $32.06\text{kg}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{SO}}_2$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 32.06\text{kg}{\text{mol}}^{-1}+\left(2\times 16.00\text{kg}{\text{mol}}^{-1}\right)\\ & =& 32.06\text{kg}{\text{mol}}^{-1}+32.00\text{kg}{\text{mol}}^{-1}\\ & =& 64.06\text{kg}{\text{mol}}^{-1}\end{array}$

Therefore, the conversion factor to obtain kilograms of ${\text{SO}}_2$ from kilomoles of ${\text{SO}}_2$ is given below.

$\frac{64.06\text{kg}{\text{SO}}_2}{\text{1}\text{kmol}{\text{SO}}_2}$

The formula to calculate the mass of ${\text{SO}}_2$ from ${\text{CS}}_2$ is given below.

$\text{Mass}\text{of}{\text{SO}}_2=\left(\begin{array}{l}\text{Given}\text{mass}\text{of}{\text{CS}}_2\\ \times \text{Conversion}\text{factor}\text{to}{\text{obtain moles of SO}}_2\\ \times \text{Conversion}\text{factor}\text{to obtain to}\text{obtain moles}\text{of}{\text{CS}}_2\\ \times \text{Conversion}\text{factor}\text{to obtain grams of}{\text{SO}}_2\end{array}\right)$…(1)

The mass of ${\text{CS}}_2$ is $\text{5}{\text{.00×10}}^{\text{2}}\text{kg}$.

Substitute the mass of ${\text{CS}}_2$ and the conversion factors in equation (1).

$\begin{array}{rll}\text{Ideal}\text{mass}\text{of}{\text{SO}}_2& =& \text{5}{\text{.00×10}}^{\text{2}}\text{kg}\times \frac{\text{2}\text{kmol}{\text{SO}}_2}{1\text{kmol}{\text{CS}}_2}\times \frac{\text{1}\text{kmol}{\text{CS}}_2}{\text{76}\text{.13}\text{kg}{\text{CS}}_2}\times \frac{64.06\text{kg}{\text{SO}}_2}{\text{1}\text{kmol}{\text{SO}}_2}\\ & =& \text{5}{\text{.00×10}}^{\text{2}}\text{kg}\times 2\times \frac{1}{\text{76}\text{.13}\text{kg}{\text{CS}}_2}\times 64.06\text{kg}{\text{SO}}_2\\ & =& 841.455\text{kg}\end{array}$

The percentage yield is calculated by the formula given below.

$\text{Percentage}\text{yield}=\frac{\text{Actual}\text{yield}}{\text{Ideal}\text{yield}}\times 100$…(2)

The percentage is $92.9\%$.

The ideal yield is $841.455\text{kg}$.

Substitute the value of ideal yield and percentage in equation (2).

$\begin{array}{rll}92.9\%& =& \frac{\text{Actual}\text{yield}}{841.455\text{kg}}\times 100\\ \text{Actual}\text{yield}& =& \frac{92.9\times 841.455\text{kg}}{100}\\ & =& 782\text{kg}\end{array}$

Therefore, the number of kilograms of sulfur dioxide that will result from burning $5.00\times {10}^2$ kilograms of carbon disulfide is $782\text{kg}$.

Conclusion

The number of kilograms of sulfur dioxide that will result from burning $5.00\times {10}^2$ kilograms of carbon disulfide is $782\text{kg}$.