2.16 gm of pure Na₂CO₃ is added to 400 ml decinormal solution of H₂SO₄. How many grams of H₂SO₄is further required to neutralize the resultant solution completely?

Balanced equation:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O

We Know:

Two substances neutralize each other when the molar ratio 1 : 1.

Molar mass of Na₂CO₃ = 106 g/mol

Amount of Na₂CO₃ taken = 2.16 g

Moles in 2.16 g Na₂CO₃ = (2.16 g) / (106 g/mol) = 0.0204 mol

Moles of H₂SO₄ needed to neutralize 0.0204 mol of Na₂CO₃ = 0.0204 mol

Concentration of H₂SO₄ solution = 0.1 N

= (0.1/2) M = 0.05 M

Moles of H₂SO₄ in 1000 mL (1 L) = 0.05

Moles of H₂SO₄ in 400 mL solution

= 0.05 × 400/1000 = 0.02 mol

Total moles of H₂SO₄ needed = 0.0204

Further amount of H₂SO₄ needed

= 0.0204 mol - 0.02 mol

= 0.0004 mol

Molar mass of H₂SO₄ = 98 g/mol

Mass of 0.0004 mol of H₂SO₄ = 98 g × 0.0004

= 0.0392 g

Mass of H₂SO₄ further needed = 0.0392 gram