Question

36 g of carbon was mixed with 142 g of chlorine to form carbon tetrachloride. Calculate the moles of product obtained and the mass of excess reactant left.

• A. 3 mol, 24 g
• B. 2 mol, 30 g
• C. 2 mol, 12 g
• D. 1 mol, 24 g

Answer 1

The correct option is D 1 mol, 24 g

$C+2C{l}_2\to CC{l}_4$
Moles of carbon $=\frac{36}{12}=3$ mol
Moles of chlorine $=\frac{142}{71}=2$ mol
Finding the limiting reagent:
$\text{For carbon}=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{3}{1}=3$
$\text{For chlorine}=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{2}{2}=1$
Hence, chlorine is the limiting reagent
As per stoichiometry,
2 moles of chlorine produce 1 mole of $CC{l}_4$
2 moles of chlorine reacts with 1 mole of carbon
Moles of excess reactant (carbon) left $=3-1=2$ mol
Mass of excess reactant left = 24 g