Question

$8\text{g}$ of $O_2$ reacts with excess of $H_2$ to form water. How much water will be formed as a result?

• A. $12\text{g}$
• B. $15\text{g}$
• C. $5\text{g}$
• D. $9\text{g}$

Answer 1

The correct option is D $9\text{g}$

Hydrogen reacts with oxygen as follows:
$2H_2+O_2\to 2H_{2}O$

Number of moles of $O_2=\frac{8}{32}=0.25\text{mol}$

As 2 mol of $H_2$ react with 1 mol of $O_2$ to form 2 mol of $H_{2}O$.
So, 0.25 mol of $O_2$ will form = $\frac{2}{1}\times 0.25=0.5\text{mol}$ of $H_{2}O$
Weight of $H_{2}O$ formed = Moles $\times$ Molar mass
$=0.5\text{mol}\times 18\text{g/mol}=9\text{g}{H}_{2}O$