Question

Chloroform $\left(CHC{l}_3\right)$ reacts with chlorine $\left(C{l}_2\right)$ to form carbon tetrachloride $\left(CC{l}_4\right)$ and hydrogen chloride $\left(HCl\right)$. In an experiment 25 grams of chloroform and 25 grams of chlorine were mixed. What is the maximum yield of $CC{l}_4$ in grams?

• A. 24.98 g
• B. 32.18 g
• C. 44.86 g
• D. 13.48 g

Answer 1

The correct option is B 32.18 g

$\underset{25g}{CHC{l}_3}+\underset{25g}{C{l}_2}\to CC{l}_4+HCl$
Number of moles of $CHC{l}_3=\frac{25g}{119.5g/mol}=0.209mol$
Number of moles of $C{l}_2=\frac{25g}{71g/mol}=0.352mol$
$CHC{l}_3$ is the limiting reagent and $C{l}_2$ is the excess reagent
So, $CC{l}_4$ formed will be $=\frac{0.209moleofCHC{l}_3\times 1moleofCC{l}_4}{1moleofCHC{l}_3}=0.209mol$
Weight of $CC{l}_4=0.209mol\times 154g/mol=32.18g$