Question

Carbon reacts with chlorine to form $CC{l}_4\cdot 36$ gm of carbon was mixed with $142$ gm of $C{l}_2$. Calculate mass of $CC{l}_4$ produced and the remaining mass of reactant.

Answer 1

$C+2C{l}_2⟶CC{l}_4$

$1$ mole of $C$ is reacting with $2$ mole of $CC{l}_4$ and gives $1$ mole of $CC{l}_4$.

Mole of $C$ in $36g=\frac{36}{12}=3$ mole

Mole of $C{l}_2$ in $142g=\frac{142}{71}=2$ mole

Thus $C{l}_2$ is limiting reagent

Mole $C+2C{l}_2⟶CC{l}_4$

At $t=0$ $3$ $2$ -

After reaction $3-1$ $2-2\times 1$ $1$

Remaining $2$ $0$ $1$

Amount of $CC{l}_4$ formed= $1\times 154g=154g$

Amount of $C$ remaining= $2\times 12g=24g$