Question

$H_2+{Cl}_2\to 2HCl$
If $2.02g$ of $H_2$ and $35.45g$ of ${Cl}_2$ are allowed to react, what is the limiting reactant and what mass of $HCl$ will be formed?

• A. $H_2,72.92g$
• B. ${Cl}_2,36.46g$
• C. $H_2,36.46g$
• D. ${Cl}_2,18.23g$

Answer 1

The correct option is B ${Cl}_2,36.46g$

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

No. of moles of Hydrogen $=\frac{2.02}{2.02}=1mole$

No. of moles of Oxygen $=\frac{35.45}{70.9}=0.5mole$

$1mole$ of $H_2$ react with $1mole$ of Chlorine to give $2moles$ of HCl. But Chlorine has only $0.5mole$. So only $1mole$ of HCl is formed.

$\Rightarrow$ Mass of $1mole$ of HCl $=1.01+35.45=36.469$

As, chlorine is completely consumed, $C{l}_2$ is limiting reagent.