Question

Which of the following statements is/are correct?

A mixture containing $64.0$ g of $H_2$ and $64.0$ g of $O_2$ is ignited, so that water is formed as follows:

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

• A. $H_2$ is the limiting reagent
• B. $O_2$ is the limiting reagent
• C. The reaction mixture contains $72.0$ g of $H_{2}O$ and $56.0$ g of unreacted $H_2$
• D. The reaction mixture contains $56.0$ g of $H_{2}O$ and $72.0$ g of unreacted $H_2$

Answer 1

Answer: (B), (C)

The correct options are
B $O_2$ is the limiting reagent
C The reaction mixture contains $72.0$ g of $H_{2}O$ and $56.0$ g of unreacted $H_2$

The number of moles of $H_2$ and $O_2$ are given below:

$n_{H_2}=\frac{64}{2}=32$ mol and $n_{O_2}=\frac{64}{32}=2$ mol

$1$ mol of $O_2$ requires $2$ mol of $H_2$.

$2$ mol of $O_2$ requires $4$ mol g $H_2$.

Since, $H_2$ is present in excess, therefore, $O_2$ is the limiting reagent.

$2$ moles of $O_2=4$ moles of $H_{2}O=4\times 18=72$ g $H_{2}O$

Moles of $H_2$ left $=32-4=28$ mol $=28\times 2=56$ g $H_2$

The reaction mixture contains $72$ g $H_{2}O$ and $56$ g $H_2$

Hence, option $B$ and $C$ are correct.