Question

In this reaction, which mass of magnesium sulphate is formed when 6g of magnesium reacts with excess sulphuric acid?
$Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}$
A. 8
B. 24
C. 30
D. 60

Answer 1

In the given reaction, one mole of magnesium reacts with one mole of sulphuric acid to give one mole of magnesium sulphate and one mole hydrogen. The number of moles is calculated by dividing the mass of the compound or element by the molecular weight or atomic weight of the compound or element.

Complete step by step answer:
Given,
Mass of magnesium is 6g.
The molecular weight of magnesium is 24.305 g/mol.
The formula to calculate the number of moles is given below.
$n = \dfrac{m}{M}$
n is number of moles.
m is given mass.
M is molecular weight
To calculate the moles of magnesium, substitute the values in the above equation.
$\Rightarrow n = \dfrac{{6g}}{{24.305g/mol}}$
$\Rightarrow n = 0.25$mol
The given reaction is shown below.
$Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}$
In this reaction one mole of magnesium reacts with one mole of sulphuric acid to form one mole of magnesium sulphate and one mole of hydrogen.
As one mole of magnesium sulphate is formed from one mole of magnesium.
So, 0.25 mole of magnesium sulphate is formed when 0.25 mole of magnesium is used.
The molecular weight of magnesium sulphate is 120 g/mol.
The mass of magnesium sulphate is calculated by substituting the values of moles of magnesium sulphate and molecular weight of magnesium sulphate in the given formula.
$\Rightarrow 0.25 = \dfrac{m}{{120g/mol}}$
$\Rightarrow m = 0.25 \times 120$
$\Rightarrow m = 30g$
Thus, the mass of magnesium sulphate formed when 6g of magnesium reacts with excess sulphuric acid is 30g.
Therefore, the correct option is C.

Note:
Don’t get confused by the reaction as one mole of magnesium is required to form one mole magnesium sulphate but we need to find the mass when 0.25 mole of magnesium is reacted. One mole is equal to the atomic mass of the element.