Chapter 11.2, Problem 16PP

Interpretation Introduction

Interpretation:

A chemical equation for the reaction of sulfur dioxide with oxygen and water has to be balanced. $\text{2}\text{.50}\text{g}$ of ${\text{SO}}_2$ reacts with excess oxygen and water. The number of grams of sulfuric acid produced is to be calculated.

Concept introduction:

The substances that undergo chemical reaction are defined as reactants and the substances thus formed are known as products.

The number of moles is calculated as shown below.

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ (1)

Answer to Problem 16PP

The balanced chemical equation is shown below.

${\text{2SO}}_2+2{\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}\to 2{\text{H}}_{\text{2}}{\text{SO}}_4$

The mass of sulfuric acid produced is $\text{3}\text{.825}\text{g}$.

Explanation of Solution

According to law of conservation of mass it is not possible to form the mass or to destroy the mass. The mass always remains conserved. The balancing of chemical equation equalizes the number of atoms of each element at the reactant side as well as at the product side. Thus, this balancing ultimately makes the mass of each species equal on both the sides of the chemical equation. The mass of reactants is always equal to mass of products in a balanced chemical equation.

The reaction of sulfur dioxide with water and oxygen to produce sulfuric acid is shown below.

${\text{2SO}}_2+2{\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}\to 2{\text{H}}_{\text{2}}{\text{SO}}_4$

The molar mass of sulfur dioxide is $\text{64}\text{.066}\text{g}/\text{mol}$. Substitute the given value of mass and molar mass in the above formula.

$\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\ =\frac{2.50\text{g}}{64.066\text{g}/\text{mol}}\\ =0.039\text{mol}\end{array}$

The balanced chemical equation shows that two moles of sulfuric acid reacts with two moles of water and one mole of oxygen to form two moles of sulfuric acid.

Therefore, the number of moles of sulfuric acid produced is,

$\begin{array}{c}2\text{mol}\text{of}{\text{SO}}_2=\text{2}\text{mol}\text{of}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\\ 1\text{mol}\text{of}{\text{SO}}_2=\frac{\text{2}}{2}\text{mol}\text{of}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\\ 0.039\text{mol}\text{of}{\text{SO}}_2=\frac{\text{2}}{2}\times 0.039\text{mol}\text{of}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\\ =0.039\text{mol}\text{of}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

The molar mass of sulfuric acid is $\text{98}\text{.079}\text{g}/\text{mol}$. Substitute the given value of mass and molar mass in the above formula.

$\begin{array}{c}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ 0.039\text{moles}=\frac{\text{Mass}}{98.079\text{g}/\text{mol}}\\ \text{Mass}=3.825\text{g}\end{array}$

Conclusion

The balanced chemical equation is shown below.

${\text{2SO}}_2+2{\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}\to 2{\text{H}}_{\text{2}}{\text{SO}}_4$

The mass of sulfuric acid produced is $\text{3}\text{.825}\text{g}$.