Chapter 12.1, Problem 9LC

Interpretation Introduction

Interpretation: The given equation needs to be interpreted in terms of relative numbers of representative particles, numbers of moles, and masses of reactants and products.

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 9LC

In a given equation, there are $2\times 6.02\times {10}^{23}\text{atoms of K }+\text{ 2}\times 6.02\times {10}^{23}{\text{molecules H}}_2\text{O}$ in the reactants and $2\times 6.02\times {10}^{23}\text{molecules KOH }+\text{ 2}\times 6.02\times {10}^{23}\text{ atoms H}$ in products. The number of moles of reactant are $2\text{ mol K }+2{\text{ mol H}}_{2}O$ and products are $2\text{ mole KOH }+1{\text{ mol H}}_2$ . The masses of reactants are: $78.2g\text{ K }+36.0g{\text{ H}}_2\text{O}$ and The mass of products are: $112.2g\text{ KOH }+2.0g{\text{ H}}_2$

Explanation of Solution

The balanced equation given is:

  $\text{2K(s)}+2{\text{H}}_2\text{O(l)}\to 2{\text{KOH(aq)+H}}_2(g)$

On the reactant side: The number of representative particles are:

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

On the product side: The number of representative particles are:

  $2\times 6.02\times {10}^{23}\text{molecules KOH }+\text{ 2}\times 6.02\times {10}^{23}\text{ atoms H}$

On the reactant side: The number of moles are:

  $2\text{ mol K }+2{\text{ mol H}}_{2}O$

On the product side: The number of moles are:

  $2\text{ mole KOH }+1{\text{ mol H}}_2$

On the reactant side: The mass of reactants are:

  $\begin{array}{l}2mol\times 39.098\frac{g}{mol}\text{ K }+\text{ 1 mol}\times 36.0\frac{g}{mol}{\text{ H}}_2\text{O} \\ =78.2gK+36.0g{\text{ H}}_2\text{O}\end{array}$

Total mass in the reactant side is $114.2g$

On the product side: The mass of products are:

  $\begin{array}{l}\text{2 mol × 56}\text{.105 }\frac{\text{g}}{\text{mol}}\text{ KOH + 1 mol × 2}\text{.0}\frac{\text{g}}{\text{mol}}\\ 112.2g\text{ KOH }+2.0g{\text{ H}}_2\end{array}$

Total mass in the product side is $114.2g$