Question

A mixture containing $100$ gm $H_2$ and $100$ gm $O_2$ is ignited so that water is formed according to the reaction,

$2H_2+O_2\to 2H_{2}O$.

How much water will be formed?

• A. $112.5$ gm
• B. $50$ gm
• C. $25$ gm
• D. $200$ gm

Answer 1

The correct option is A $112.5$ gm

$2H_2+O_2\to 2H_{2}O$

100 gm $H_2=\frac{100}{2}=50$ mole

100 gm $O_2=\frac{100}{32}=\frac{25}{8}$ mole

According to stiochiometry

$2H_2+O_2\to 2H_{2}O$

Since oxygen is limiting agent

$H_{2}O$ mole of $H_{2}O=\frac{25}{4}$

Weight of $H_{2}O=\frac{25}{4}\times 18=112.5gm$