Question

Carbon reacts with chlorine to form ${CCl}_4$. $36$ g of carbon was mixed with $142$ g of ${Cl}_2$. Calculate the mass of ${CCl}_4$ produced and the remaining mass of reactant.

• A. $W_{\left(CC{l}_4\right)}=154$ g, $W_c=24$ g
• B. $W_{\left(CC{l}_4\right)}=156$ g, $W_c=24$ g
• C. $W_{\left(CC{l}_4\right)}=56$ g, $W_c=2$ g
• D. $W_{\left({CCl}_4\right)}=26$ g, $W_c=54$ g

Answer 1

The correct option is A $W_{\left(CC{l}_4\right)}=154$ g, $W_c=24$ g

The balanced chemical reaction is $C+2C{l}_2\to CC{l}_4$.

$36$ g of carbon will react with $\frac{36}{12}\times 2\times 71=426$ g of chlorine.

However, only $142$ g of chlorine is present.

Thus, chlorine is the limiting reagent.

$142$ g of chlorine will produce $\frac{142\times 154}{71\times 2}=154$ g of carbon tetrachloride.

$142$ g ($2$ moles) of chlorine will react with $12$ g ($1$ mole) of carbon.

The amount of carbon remaining unreacted after the reaction is $36-12=24$ g.