Question

In a reaction vessel, 100 g $H_2$ and 100 g $C{l}_2$ are inbred and suitable conditions are provided for take following reaction:

$H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right)$

The amount of excess reactant remaining is:

• A. 50 g
• B. 97.2 g
• C. 46 g
• D. 64 g

Answer 1

The correct option is B 97.2 g

100 g of $H_2$ = 50 mole

100 g of $C{l}_2$ = 1.4 mole

According to the reaction 1 mole of $H_2$ is reacting with 1 mole of $C{l}_2$

So, $C{l}_2$ is th limiting reagent

and amount of $H_2$ consumed = 1.4 moles

So, amount of excess reactant = 50 - 1.4 = 48.6 moles

Amount of $H_2$ left = 48.6 \times 2 = 97.2 g

Hence, the correct option is $\left(B\right)$.