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 $HCI$ formed (at 90% yield) will be:

• A. 36.8 g
• B. 62.5 g
• C. 80 g
• D. 91.98 g

Answer 1

The correct option is D 91.98 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

So, moles of $HCl$ formed = 2 $\times$ 1.4 mole

Amount of $HCl$ = 2.8 $\times$ 36.5 = 102.2 g

Since, the reaction is giving 90 % yield

therefore , amount of $HCl$ formed = $102.2\times \frac{90}{100}$

Amount of HCl formed = 91.98 g

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