Question

If 241.5 g of zinc sulphate $\left(ZnS{O}_4\right)$ reacts with 81 g of aluminium to give aluminium sulphate. Calculate the mass of the excess reagent left.

$3ZnS{O}_4+2Al\to A{l}_2(S{O}_4{)}_3+3Zn$
(take molar mass of $Zn$ as 65 g)

• A. $27g$
• B. $54g$
• C. $81g$
• D. $161g$

Answer 1

The correct option is B $54g$

Given balanced chemical reaction is:
$3ZnS{O}_4+2Al\to A{l}_2(S{O}_4{)}_3+3Zn$

The molar mass of $ZnS{O}_4=65+32+64=161g/mol$

$\text{So, 241.5 g}ZnS{O}_4=\frac{241.5}{161}mol=1.5molZnS{O}_4$
$\text{Again, 81 g}Al=\frac{81}{27}mol=3molAl$

From the balanced chemical equation,
3 moles of $ZnS{O}_4$ reacts with 2 moles of $Al$.

Thus, 1.5 moles of $ZnS{O}_4$ will react with 1 mole of $Al$.

We are given 3 moles of Al.

Thus, the excess reagent here is $Al$.
$\text{Amount of excess}Al=3-1mol=2\times 27g=54g$