Question

In an experiment, $4g$ of $H_2$ reacts with $32g$ of $O_2$ to produce $36g$ of $H_{2}O$. Calculate the amount of $O_2$ required to react with $12g$ of $H_2$ to produce $108g$ of $H_{2}O$.

• A. $16g$
• B. $32g$
• C. $48g$
• D. $96g$

Answer 1

The correct option is D $96g$

The reaction of $H_2$ and $O_2$ to produce $H_{2}O$ is as follows:
$2H_2+O_2\to 2H_{2}O$
Given that $4g$ of $H_2$ reacts with $32g$ of $O_2$ to produce $36g$ of $H_{2}O$.
According to the law of conservation of mass,
Total mass of reactants = Total mass of products

	$\text{Mass of}{H}_2$	$\text{Mass of}{O}_2$	$\text{Mass of}{H}_{2}O$
$\text{Condition}1$	$4g$	$32g$	$36g$
$\text{Condition}2$	$12g$	$?$	$108g$

We have,
$\left(\text{Mass of}{H}_2\right)+\left(\text{Mass of}{O}_2\right)=\left(\text{Mass of}{H}_{2}O\right)$
$\Rightarrow 12g+\left(\text{Mass of}{O}_2\right)=108g$
$\Rightarrow \text{Mass of}{O}_2=(108-12)g$
$\Rightarrow \text{Mass of}{O}_2=96g$