Q3.133CP - Chemistry: Molecular Nature Of Matter And Change

Question

Hydrogen gas has been suggested as a clean fuel because it produces only water vapour when it burns.Ifthe reaction has a 98.8% yield, what mass of hydrogen forms 105 kg of water?

Step-by-Step Solution

Answer

TheMass of hydrogen formed from of water is $1.19\times {10}^4\text{g}$

Step 1: Calculation of theoretical yield of the water vapors:

Given:

Percentage yield of water vapor =$98.8\%$

Mass of water =$\text{105kg}$

$$\begin{gathered} \text{Percentage yield=}\frac{\text{Actual yield of water}}{\text{Theoretical yield of water}}\text{×}100 \\ \text{Theoretical yield of water = }\frac{\text{Actual yield of water}}{\text{Percentage yield of water}}\times 100 \\ \text{Theoretical yield of water =}\frac{{\displaystyle 105\text{kg}}}{98.8\%}\text{×100} \\ =106\text{ kg} \end{gathered}$$

Step 2: Calculation of Hydrogen gas produced.

Let us calculate the mass of hydrogen gas from a theoretical yield of water.

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

Molar mass of ${\text{H}}_2\text{O = 18.02 g/mol}$

Molar mass of ${\text{H}}_2\text{ = 2.02 g/mol}$

${\text{1 kg = 10}}^{\text{3}}\text{g}$

$\text{106 kg = 106 }\times {\text{ 10}}^3\text{g}$

${\text{Mass of H}}_2\text{ = 106 }\times {\text{ 10}}^3\text{g }\times \frac{1{\text{ mol of H}}_2\text{O }}{18.02{\text{ g of H}}_2\text{O}}\times \frac{2{\text{mol of H}}_2}{2{\text{mol of H}}_2\text{O }}\times \frac{2.02{\text{g of H}}_2}{{\text{1mol of H}}_2}$

${\text{Mass of H}}_{\text{2}}{\text{=1.19×10}}^{\text{4}}\text{g}$

Result:

The mass of hydrogen gas produced is${\text{1.19×10}}^{\text{4}}\text{g}$