Chapter 11.1, Problem 9SSC

Interpretation Introduction

Interpretation:

Water and oxygen are produced on decomposition of hydrogen peroxide. The balanced chemical equation and possible mole ratios for the reaction has to determined.

Concept introduction:

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

Answer to Problem 9SSC

The balanced chemical equation is shown below:

${\text{2H}}_2{\text{O}}_2\left(l\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)$

The mole ratio for given chemical equation is shown below:

$\begin{array}{l}\frac{2{\text{mol H}}_2{\text{O}}_2}{2\text{mol}{\text{H}}_2\text{O}}\text{ and }\frac{2{\text{mol H}}_2{\text{O}}_2}{1\text{mol}{\text{O}}_2}\\ \frac{2{\text{mol H}}_2\text{O}}{2\text{mol}{\text{H}}_2{\text{O}}_2}\text{ and }\frac{2{\text{mol H}}_2\text{O}}{1\text{mol}{\text{O}}_2}\\ \frac{1{\text{mol O}}_2}{2\text{mol}{\text{H}}_2{\text{O}}_2}\text{ and }\frac{1\text{mol}{\text{O}}_2}{2{\text{mol H}}_2\text{O}}\end{array}$

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 balanced chemical equation is shown below:

${\text{2H}}_2{\text{O}}_2\left(l\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)$

The mole ratio for given chemical equation is shown below:

$\begin{array}{l}\frac{2{\text{mol H}}_2{\text{O}}_2}{2\text{mol}{\text{H}}_2\text{O}}\text{ and }\frac{2{\text{mol H}}_2{\text{O}}_2}{1\text{mol}{\text{O}}_2}\\ \frac{2{\text{mol H}}_2\text{O}}{2\text{mol}{\text{H}}_2{\text{O}}_2}\text{ and }\frac{2{\text{mol H}}_2\text{O}}{1\text{mol}{\text{O}}_2}\\ \frac{1{\text{mol O}}_2}{2\text{mol}{\text{H}}_2{\text{O}}_2}\text{ and }\frac{1\text{mol}{\text{O}}_2}{2{\text{mol H}}_2\text{O}}\end{array}$

Conclusion

The balanced chemical equation is shown below:

${\text{2H}}_2{\text{O}}_2\left(l\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)$

The mole ratio for given chemical equation is shown below:

$\begin{array}{l}\frac{2{\text{mol H}}_2{\text{O}}_2}{2\text{mol}{\text{H}}_2\text{O}}\text{ and }\frac{2{\text{mol H}}_2{\text{O}}_2}{1\text{mol}{\text{O}}_2}\\ \frac{2{\text{mol H}}_2\text{O}}{2\text{mol}{\text{H}}_2{\text{O}}_2}\text{ and }\frac{2{\text{mol H}}_2\text{O}}{1\text{mol}{\text{O}}_2}\\ \frac{1{\text{mol O}}_2}{2\text{mol}{\text{H}}_2{\text{O}}_2}\text{ and }\frac{1\text{mol}{\text{O}}_2}{2{\text{mol H}}_2\text{O}}\end{array}$