Question

$H_2\left(g\right)+{Cl}_2\left(g\right)\to 2HCl\left(g\right)$
If $70.9g$ of chlorine reacts withe excess hydrogen, what volume of $HCl$ will be formed? Assume STP conditions.

• A. $11.2L$
• B. $22.4L$
• C. $5.6L$
• D. $44.8L$

Answer 1

The correct option is D $44.8L$

At STP conditions, $1mole$ of gas occupies $22.4Litres$

Given, $H_2+C{l}_2\to 2HCl$

$\ast 1mole$ of Chlorine produce $2moles$ of HCl

$70.9g$ of Chlorine $=\frac{70.9g}{70.9\left(\text{Molar mass of Chlorine}\right)}=1mole$ of Chlorine.

So, $70.9g$ Chlorine produce $2mole$ of HCl i.e. $2\times 22.4=44.8L$ of volume.