Question

$Mg+H_{2}S{O}_4\to MgS{O}_4+H_2$
In this reaction, which mass of magnesium sulphate is formed when 6g of magnesium react with excess sulphuric acid?

• A. $8$
• B. $24$
• C. $30$
• D. $60$

Answer 1

Answer: (C)

$30$

Solution:- (C) $30$

As we know that,

$\text{No. of moles}=\frac{\text{wt.}}{\text{mol. wt.}}$

Now,

Wt. of $Mg=6g$

Mol. wt. of $Mg=24g$

No. of moles of $Mg$ in $6g=\frac{6}{24}=0.25\text{moles}$

$Mg+H_{2}S{O}_4⟶MgS{O}_4+H_2$

from the above reaction,

$1$ mole of magnesium sulphate is formed when $1$ mole of magnesium reacts.

Therefore,

$0.25$ mole of magnesium sulphate is formed when $0.25$ mole of magnesium reacts.

Now,

Mol. wt. of $MgS{O}_4=120g$

Therefore,

Wt. of $0.25$ moles of $MgS{O}_4=0.25\times 120=30g$

Hence $30g$ of magnesium sulphate is formd when $6g$ of magnesium reacts with excess of $H_{2}S{O}_4$.