Question

$10g$ of carbon reacts with $100gC{l_2}$ to form $CC{l_4}$. The correct statement is-
a.Carbon is a limiting reagent
b.$C{l_2}$ is a limiting reagent
c.$107.8g$ $CC{l_4}$ is formed
d.$0.833$ moles of $CC{l_4}$ are formed

Answer 1

Hint: First of all, we need to write down the reaction taking place between carbon and chlorine to form carbon tetrachloride. After that we will be able to find out the amount of formation of carbon tetrachloride.

Complete step by step answer:
In the question, it is given that carbon and chlorine are reacting together to form carbon tetrachloride. Therefore, we can write down the reaction taking place in between them:
$C + 2C{l_2} \to CC{l_4}$
So, from the above equation, we can say that one mole of carbon is reacting with two moles of chlorine to give carbon tetrachloride, which means that $12g$ of carbon is reacting with $2 \times 71 = 142g$ because we know that the atomic weight of carbon is $12g$and of chlorine is $35.5g$ and here two moles are reacting so it becomes $71g$. So, we know that $12g$ of carbon requires $142g$ of chlorine, so by using the unitary method, we will find out the amount of chlorine required to react with $10g$ of carbon.
$10g$ of carbon $ = \dfrac{{2 \times 71 \times 10}}{{12}} = 118.33g$ of $C{l_2}$ is required
But the available chlorine is $100g$ which is less than the required chlorine, that is, $118.33g$
Therefore, we can say that chlorine is a limiting reagent.
Now, the further calculations will be done with the available amount of chlorine which is $100g$.
Let’s find out the amount of $CC{l_4}$ produced. We know that two moles of chlorine and one mole of carbon is required to produce one mole of $CC{l_4}$, so, if we find out the weight of $CC{l_4}$ in one mole, than that will be our amount of $CC{l_4}$ produced.
$\therefore $ $12 + 4 \times 35.5 = 154grams$ of $CC{l_4}$ is produced
Now, let’s find out the amount of $CC{l_4}$ produced in $100g$ of chlorine:
$100g$ of $C{l_2}$ $ = \dfrac{{154 \times 100}}{{2 \times 71}} = 108.45grams$ $CC{l_4}$

So, the correct options are option b) $C{l_2}$ is a limiting reagent and option c) $107.8g$ $CC{l_4}$ is formed.

Note:
After writing the equations, we should always keep in mind to balance the equations because then only we will be able to find out how many moles of the reagents are required to react together and then only we will be able to find out the exact amount of product formed.