Chapter 11, Problem 67A

Interpretation Introduction

Interpretation:

The mass of methane that is required to produce 50.0 g of ${\text{CHCl}}_{\text{3}}$ for the given balance chemical equation needs to be determined.

Concept introduction:

The mole concept provides the relation between moles, mass and molar mass. The mathematical expression of mole concept can write as:

$\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}$

1 mole of any substance contains $\text{6}\text{.022 }\times {\text{ 10}}^{\text{23}}$ number of particles or atoms or molecules.

Answer to Problem 67A

$\text{6}{\text{.70 g CH}}_{\text{4}}$

Explanation of Solution

Given information:

${\text{CH}}_{\text{4}}{\text{(g)+ 3 Cl}}_{\text{2(g)}}\stackrel{}{\to }{\text{CHCl}}_{\text{3}}\text{(g) + 3 HCl(g)}$

The given reaction:

${\text{CH}}_{\text{4}}{\text{(g)+ 3 Cl}}_{\text{2(g)}}\stackrel{}{\to }{\text{CHCl}}_{\text{3}}\text{(g) + 3 HCl(g)}$

Mass of ${\text{CHCl}}_{\text{3}}$ = 50.0 g

Molar mass of ${\text{CHCl}}_{\text{3}}$ = 119.4 g/mol

Molar mass of ${\text{CH}}_{\text{4}}$ = 16.0 g/mol

The mole ratio relates the moles of reactant with other reactant and product molecules in the balance chemical equation.

1 mole of ${\text{CH}}_{\text{4}}$ = 1 mole of ${\text{CHCl}}_{\text{3}}$

Calculate mass of ${\text{CH}}_{\text{4}}$ with the help of mole ratio:

$\text{50}{\text{.0 g CHCl}}_{\text{3}}\text{ ×}\frac{{\text{1 mole CHCl}}_{\text{3}}}{\text{119}{\text{.4 g CHCl}}_{\text{3}}}\text{×}\frac{{\text{1 mole CH}}_{\text{4}}}{{\text{1 mole CHCl}}_{\text{3}}}\text{×}\frac{\text{16}{\text{.0 g CH}}_{\text{4}}}{{\text{1 mole CH}}_{\text{4}}}\text{= 6}{\text{.70 g CH}}_{\text{4}}$

Conclusion

Mole ratio is used to calculate the mass of product.