Question

When 10 ml of $H_2$ and 12.5 ml of $C{l}_2$ are allowed to react,the final mixture contains under the same conditions.

• A. 22.5 ml of $HCl$
• B. 12.5 ml of $HCl$
• C. 20 ml of $HCl$
• D. 2.5 ml of $C{l}_2$ and 20 ml $HCl$

Answer 1

The correct option is A 22.5 ml of $HCl$

$10ml{H}_2$ and $1.25mlC{l}_2$

and 0.45 and 0.55 millimoles of $H_2$ and $C{l}_2$

The reaction is $H_2+C{l}_2\to 2HCl$.

thus, 1 moles of $H_2$ reacts with 1 mole of $C{l}_2$ to from two of $HCl.$

0.45 mililmoles of $H_2$ will react 0.45 millmoles of $C{l}_2$ to from

0.9 millimoles of $HCl$

0.1 millimoles of $C{l}_2$ will remain unreacted.

So, the reaction mixture will contain 0.01 millimoles of $C{l}_2$ and 0.9 millimoles of $HCl$

$20.16$ mL of $HCL$ and $2.24mL$ of $C{l}_2$

Hence,

$2.5ml$ of $C{l}_2$ and $20ml$ $HCl$

Option $\left(D\right)$ is correct answer