Question

When magnesium metal is added to water, hydrogen gas, ${\text{H}}_2$ is released. If 72 g of hydrogen gas is liberated in this reaction, how many number of moles of hydrogen gas is produced?
$\text{Mg}$ (s) + $\text{2}$${\text{H}}_2\text{O}$ (l) $\to$ $\text{Mg}{\text{(OH)}}_2$ (aq) + ${\text{H}}_2$ (g) ​

Answer 1

Given mass of hydrogen (${\text{H}}_2)$ = 72 g
Gram molecular mass of ${\text{H}}_2$ = 2 g

Number of moles = $\frac{\text{Given mass}}{\text{Gram molecular mass}}$ [1 Mark]
$\therefore$ Number of moles of ${\text{H}}_2$ = $\frac{72}{\text{2}}$ = 32 mol
[1 Mark]