Question

The volume of Ammonia gas at STP required to neutralize 300ml of 1M H2SO4 solution is​

Answer 1

Explanation:

The volume of Ammonia gas at STP required to neutralize 300ml of 1M H2SO4 solution is

Answer 2

Given: 300mL of 1M H₂SO₄ solution.

To find: The volume of ammonia gas at STP required to neutralize the H₂SO₄ solution.

Solution:

• The reaction between ammonia and sulfuric acid is as follows,

        $2NH_{3} + H_{2}SO_{4} -> (NH_{4})_{2}SO_{4}$

• The number of moles in 300mL 1M sulphuric acid solution is calculated as,

        $\frac{300}{1000} * 1 = 0.3 moles$

• According to the equation of the reaction, one mole of sulphuric acid is neutralized by two moles of ammonia.
• So, 0.3 moles of sulfuric acid will be neutralized by,

        $\frac{0.3 * 2}{1} = 0.6 moles$

• The volume of one mole of ammonia at STP will be 22400 mL.
• So, the volume of 0.6 moles of ammonia would be equal to,

        $0.6 * 22400 = 13440 mL$

Therefore, 13440 mL of ammonia gas at STP is required to neutralize the H₂SO₄ solution.