Question

$Mg\text{+}{{\text{H}}_{2}}S{{O}_{4}}\to MgS{{O}_{4}}+{{H}_{2}}$
In this reaction which mass of magnesium sulphate is formed when 6 g of magnesium reacts with excess Sulphuric acid?
(A) 8
(B) 24
(C) 30
(D) 60

Answer 1

To calculate the amount of magnesium sulphate we have to calculate the moles of magnesium sulphate formed when one mole of magnesium reacts with Sulphuric acid.

Complete step by step solution:
As we know that number of moles is equal to the ratio of a given mass to molecular mass.
According to the question-
Given the mass of magnesium is = 6 g
And we know that molecular mass of magnesium is = 24 g
$\text{number of moles = }\dfrac{\text{given mass}}{\text{molar mass}}$
Number of moles= $\dfrac{6}{24}=0.25\text{moles}$
The number of moles of magnesium is 0.25 moles.
Now from the given equation
$Mg\text{+}{{\text{H}}_{2}}S{{O}_{4}}\to MgS{{O}_{4}}+{{H}_{2}}$
We can say that,
1 mole of magnesium reacts to form 1 mole of magnesium sulphate, so it means that 0.25 mole of magnesium will react to form 0.25 mole of magnesium sulphate.
Now we know that the molecular mass of magnesium sulphate= 120 g
And the number of moles of magnesium sulphate = 0.25 moles
Applying the formulae,
A number of moles are equal to the ratio of a given mass to molecular mass. So if we have to calculate the given mass the formula will be the product of molar mass and number of moles.
Given mass = $\left( molar\text{ mass} \right)\left( \text{number of moles} \right)$
Given mass = $\left( 120 \right)\left( 0.25 \right)$
Given mass = 30g
Hence 30g of magnesium sulphate is formed when 6 grams of magnesium reacts with excess Sulphuric acid.

Hence, the correct option is (C).

Note: Be careful while doing calculations and use the correct formula. Sulphuric acid is present in excess which means that Sulphuric acid does not determine the rate of the reaction.