Question

2.4kg of carbon is made to react with 1.35kg of aluminium to form \[{\text{A}}{{\text{l}}_4}{{\text{C}}_3}\] . The maximum amount in Kg of aluminium carbide formed is
A.\[5.4\]
B.\[3.75\]
C.\[1.05\]
D.\[1.8\]

Answer 1

Hint:In case, if a different amount of reactants is given then, we will proceed with respect to limiting reagent. Limiting reagent is the reagent which is consumed in the reaction completely. We shall write the reaction equation and use the stoichiometric ratios to find the limiting reagent and thus the amount of product formed.
Formula used: ${\text{moles = }}\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$

Complete step by step answer:
As given in question aluminium and carbon are made to react to form aluminium carbide. The balanced chemical reaction will be:
\[{\text{4Al}} + {\text{3C}} \to {\text{A}}{{\text{l}}_4}{{\text{C}}_3}\]
But in some cases one mole of reactant get finished or completely used while the other reactant is still left. Such reactants which limit the feasibility of the reaction are known as limiting reagent. First of all we need to calculate the number of moles in order to find the limiting reagent.
The molar mass of aluminium is 27. Number of moles of aluminium will be:
\[\dfrac{{1.35 \times {{10}^3}{\text{ Kg}}}}{{27}} = 50{\text{mol}}\]
The molar mass of carbon is 12. Number of moles of carbon will be:
\[\dfrac{{2.4 \times {{10}^3}{\text{ Kg}}}}{{12}} = 200{\text{mol}}\]
As we can see that there is less number of moles of aluminium and so aluminium will get finished first and limit reaction. Hence aluminium will act as limiting reagent and the product will form according to this. Now according to the balance chemical equation 4 moles of aluminium are required to form one mole of aluminium carbide.
Hence, one mole of aluminium will form \[\dfrac{1}{4}\] moles of aluminium carbide.
50 moles of aluminium will form \[\dfrac{1}{4} \times 50 = 1.25{\text{ mol}}\] of aluminium carbide.
Mass of aluminium carbide can be calculated by multiplying the number of moles with molar mass as: \[1.25 \times 144 = 1.8{\text{Kg}}\]

Hence, the correct option is D.

Note:
To find out the limiting reagent we should always calculate or deal in the number of moles of reactants and not the mass of the reactants, because number of moles is a concentration term but masses not a concentration term.