Chapter 9.2, Problem 6RQ

Interpretation Introduction

Interpretation: The mass of carbon monoxide and hydrogen gas required to form $6.0\text{ kg}$ of methanol needs to be calculated.

Concept introduction: The ratio of mass of substance to its molar mass is said to be number of moles of that substance.

Answer to Problem 6RQ

  $\text{Mass of CO = 5}\text{.245 kg}$ and ${\text{Mass of H}}_{\text{2}}\text{ = 0}\text{.755 kg}$ .

Explanation of Solution

The balanced complete chemical equation for the formation of methanol from carbon monoxide and hydrogen gas is:

  ${\text{CO(g) + 2H}}_{\text{2}}\text{(g) }\to {\text{CH}}_3\text{OH(g)}$

The number of moles of methanol is calculated as:

  $\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

The molar mass of methanol is 32.04 g/mol. So, the number of moles of methanol is:

  $\begin{array}{l}{\text{n}}_{{\text{CH}}_3\text{OH}}=\frac{6.0\times {10}^3\text{ g}}{\text{32}\text{.04 g/mol}}\\ {\text{n}}_{{\text{CH}}_3\text{OH}}=187.27\text{ mol}\end{array}$

From the balanced chemical reaction, the mole ratio of methanol and carbon monoxide is 1:1 and the mole ratio of methanol and hydrogen is 1:2 so, the number of moles of carbon dioxide and hydrogen required for the production of $187.27\text{ mol}$ methanol is:

  $187.27{\text{ mol of CH}}_3\text{OH×}\frac{\text{1 mol of CO}}{{\text{1 mol of CH}}_3\text{OH}}\text{ = }187.27\text{ mol of CO}$

  $187.27{\text{ mol of CH}}_3\text{OH×}\frac{{\text{2 mol of H}}_2}{{\text{1 mol of CH}}_3\text{OH}}\text{ = }374.54{\text{ mol of H}}_2$

Now, the mass of carbon monoxide and hydrogen required for the production of $6.0\text{ kg}$ ofmethanol is calculated as:

The molar mass of carbon monoxide and hydrogen is 28.01 g/mol and 2.016 g/mol respectively. So,

  $\begin{array}{l}\text{Number of moles of CO =}\frac{\text{Mass of CO}}{\text{Molar mass of CO}}\\ \text{187}\text{.27 mol =}\frac{\text{Mass of CO}}{\text{28}\text{.01 g/mol}}\\ \text{Mass of CO = 187}\text{.27 mol ×28}\text{.01 g/mol}\\ \text{Mass of CO = 5245}\text{.43 g}\\ \text{Mass of CO = 5}\text{.245 kg}\end{array}$

  $\begin{array}{l}{\text{Number of moles of H}}_{\text{2}}\text{ =}\frac{{\text{Mass of H}}_{\text{2}}}{{\text{Molar mass of H}}_{\text{2}}}\\ \text{374}\text{.54 mol =}\frac{{\text{Mass of H}}_{\text{2}}}{\text{2}\text{.016 g/mol}}\\ {\text{Mass of H}}_{\text{2}}\text{= 374}\text{.54 mol ×2}\text{.016 g/mol}\\ {\text{Mass of H}}_{\text{2}}\text{ = 755}\text{.07 g}\\ {\text{Mass of H}}_{\text{2}}\text{ = 0}\text{.755 kg}\end{array}$

Hence, the $\text{mass of CO = 5}\text{.245 kg}$ and ${\text{mass of H}}_{\text{2}}\text{ = 0}\text{.755 kg}$ .