Question

10 g of carbon is reacted with 100 g of $C{l}_2$ to form $CC{l}_4$. The correct statement is:

• A. Carbon is the limiting reagent
• B. Chlorine is the excess reagent
• C. 108.4 g of $CC{l}_4$ is formed
• D. 0.833 moles of $CC{l}_4$ are formed

Answer 1

The correct option is C 108.4 g of $CC{l}_4$ is formed

$C+2C{l}_2\to CC{l}_4$
Moles of carbon =$\frac{10}{12}$ = 0.833 moles
Moles of chlorine = $\frac{100}{71}=1.408$ moles
Finding the limiting reagent:
$C=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{0.833}{1}=0.833C{l}_2=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{1.408}{2}=0.704$
Hence, the limiting reagent will be chlorine.
2 moles of $C{l}_2$ produce 1 mole of $CC{l}_4$
1.408 moles produce $\frac{1}{2}\times 1.408$ = 0.704 moles of $CC{l}_4$
Mass of $CC{l}_4$ produced $=0.704\times 154=108.42$ g