Question

Carbon reacts with chlorine to form $CC{l_4}$. $36g$ of carbon was mixed with $142g$ of $C{l_2}$. Calculate the mass of $CC{l_4}$ 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) ${W_{\left( {CC{l_4}} \right)}} = 26$ g, ${W_c} = 54$ g

Answer 1

Write down the equation for the reaction between carbon and chlorine to form carbon tetrachloride. From this reaction we can calculate the weight of carbon and chlorine required to produce one mole of carbon tetrachloride.

Complete step-by-step solution: It is given to us that Carbon and Chlorine react together to form carbon tetrachloride i.e. $CC{l_4}$
We can write the equation for this reaction as follows.
$C + 2C{l_2} \to CC{l_4}$
This shows that one mole of Carbon reacts with one mole of chlorine meaning that $12$ grams of carbon reacts with $2 \times 71 = 142$ grams of chlorine.
It is also given to us that $36g$ of carbon is mixed with $142g$ of $C{l_2}$ but we have already calculated that $142$ grams of chlorine only requires $12$grams of carbon.
Hence, the amount of carbon remaining as excess is $36 - 12 = 24$ grams.
Let us now calculate the amount of carbon tetrachloride produced. In the above equation, we see that one mole of carbon and two moles of chlorine produce one mole of carbon tetrachloride. Hence, the weight of carbon tetrachloride present in one mole is the amount of carbon tetrachloride produced.
The weight of carbon tetrachloride in one mole is $12 + 4 \times 35.5 = 154$ grams.
Hence, $154$ grams of carbon tetrachloride is formed.

Hence the correct option is (A).

Note: One should always balance a reaction equation. The ratio of the weight of a compound to its gram molecular weight gives the total number of moles of that compound used. If one mole of a compound is used then its weight is equal to its gram molecular weight.