Chapter 12.1, Problem 3SP

Interpretation Introduction

Interpretation: The given equation for the formation of water needs to be interpreted in terms of the numbers of molecules and moles, and the volume of gases at STP.

Concept introduction: A mole is a SI unit of the amount of substance. In 1 mole of a substance, there are $6.022\times {10}^{23}$ particles/ions/molecules.

Answer to Problem 3SP

In a given equation, there are $4\times 6.02\times {10}^{23}\text{atoms of H }+\text{ 2}\times 6.02\times {10}^{23}\text{atoms of O}$ in the reactant and $2\times 6.02\times {10}^{23}{\text{molecules of H}}_2\text{O}$ in the product. The number of moles of reactant are ${\text{2 mol H}}_2+\text{ 1 mol}{O}_2$ and of products are $2{\text{ mole H}}_2\text{O}$ . The volume of gas in reactant are: $2\times 22.4L{\text{ of H}}_2+22.4L of{\text{ O}}_2$ and the volume of gas in products are: $\text{44}{\text{.8 L H}}_2\text{O}$

Explanation of Solution

The balanced equation given is:

  ${\text{2H}}_2\text{(}g\text{)}+{\text{O}}_2\text{(}g\text{)}\to 2{\text{H}}_2\text{O}(g\text{)}$

On the reactant side: The number of molecules are:

  $4\times 6.02\times {10}^{23}\text{atoms of H }+\text{ 2}\times 6.02\times {10}^{23}\text{atoms of O}$

On the product side: The number of molecules are:

  $2\times 6.02\times {10}^{23}{\text{molecules of H}}_2\text{O}$

On the reactant side: The number of moles are:

  ${\text{2 mol H}}_2+\text{ 1 mol}{O}_2$

On the product side: The number of moles are:

  $2{\text{ mole H}}_2\text{O}$

On the reactant side: The volume of gases at STP are:

At STP, one mole of a substance occupies $22.4L$ .

  $2\times 22.4L{\text{ of H}}_2+22.4L of{\text{ O}}_2$

On the product side: The volume of gases at STP are:

  $2\times 22.4L{\text{ of H}}_2\text{O = 44}{\text{.8 L H}}_2\text{O}$