Question

What is the maximum amount of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$ which could be formed from the reaction of ${\text{12}}{\text{.71}}$ g of Al and ${\text{10}}{\text{.09}}$ mol of ${\text{CuS}}{{\text{O}}_{\text{4}}}$?
$....{\text{Al + }}....{\text{CuS}}{{\text{O}}_{\text{4}}} \to ....{\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}{\text{ + }}....{\text{Cu}}$

Answer 1

In the above question, we have to first balance the chemical equation. Then since we can find out the number of moles of Al, we can find the limiting reagent. According to the limiting reagent, we can find the maximum amount of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$ formed.

Formula Used-
${\text{n = }}\dfrac{{\text{m}}}{{\text{M}}}$
where n= number of moles of the substance
m=given mass of the substance
M=molar mass of the substance.

Complete step-by-step answer: Let us first the equation:
$....{\text{Al + }}....{\text{CuS}}{{\text{O}}_{\text{4}}} \to ....{\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}{\text{ + }}....{\text{Cu}}$
2 Al is present on the product side and only one on the reactant side. So, let us first balance it.
${\text{2Al + CuS}}{{\text{O}}_{\text{4}}} \to {\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}{\text{ + Cu}}$
3 molecules of ${\text{S}}{{\text{O}}_{\text{4}}}$ are present on the product side and one on the reactant side. Now, let us balance both ${\text{S}}{{\text{O}}_{\text{4}}}$ and Cu.
${\text{2Al + 3CuS}}{{\text{O}}_{\text{4}}} \to {\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}{\text{ + 3Cu}}$
This implies that 2 mole of Al reacts with 3 mole of ${\text{CuS}}{{\text{O}}_{\text{4}}}$ to form 1 mole of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$.
Given mass of Al = ${\text{12}}{\text{.71}}$g
Molar mass of Al= 27g
${\text{n = }}\dfrac{{\text{m}}}{{\text{M}}}{\text{ = }}\dfrac{{{\text{12}}{\text{.71}}}}{{{\text{27}}}}{\text{ = 0}}{\text{.47}}$
Since, the amount of ${\text{CuS}}{{\text{O}}_{\text{4}}}$present is excess (${\text{10}}{\text{.09}}$mol). So, the limiting reagent is Al.
2 moles of Al gives 1 mole of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$.
So, ${\text{0}}{\text{.47}}$moles of Al gives $\dfrac{{\text{1}}}{{\text{2}}}{{ \times 0}}{\text{.47 = 0}}{\text{.235}}$ moles of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$.
Molar mass of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$= 2${{ \times }}$ atomic mass of Al + 3 ${{ \times }}$ atomic mass of S + 12 ${{ \times }}$ atomic mass of oxygen
= ${{2 \times 27 + 3 \times 32 + 12 \times 16}}$ = ${\text{54 + 96 + 192 = 342}}$
We know that ${\text{n = }}\dfrac{{\text{m}}}{{\text{M}}}$
So, rearranging, we get: ${{m = n \times M = 0}}{{.235 \times 342 = 80}}{\text{.37}}$g
Therefore, the weight of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$ formed is ${\text{80}}{\text{.37}}$g.

Note: In these types of questions where mass and moles or any other information is available for all the reactants then we have to first find the limiting reagent and then proceed accordingly.