Data

Volume = 200cm³ (0.2dm³) 0f H₂SO₂

Molar volume of S.T.P = 22.4dm³/mole
Asked:
i. Mass of Barium Sulphate
ii. Moles of Barium Sulphate

Balanced chemical equation for the reaction.
Ba(NO₃)₂(aq) + H₂SO₄(aq) → BaSO₄(S) + 2HNO₃(aq)

1 mole: 1 mole: 1 mole: 2 mole

i/ Mass of Barium sulphate formed

convert 0.2dm³ into mole
Mole of H₂ = 0.2dm³ / 22.4dm³
= 0.00892 mole.

write mole equivalent between H₂ and BaSO4
1 Mole of H₂ Ξ 1 mole of BaSO4
0.008928 mole Ξ ?? of BaSO4

= (1 mole × 0.008928 mole) / 1 mole
= 0.008928 mole of BaSO4

Mass of BaSO4 = Mole × Molar mass
Molar mass of BaSO4 = 56 +32 +(16×4)
= 56 +32 +64
= 152g/mole

Mass = 0.008928 mole × 152g/mole
= 1.357056g

1.357056g of Barium Sulphate was produced.

ii/calculate the mole of Barium Sulphate formed.

From equivalent moles between H₂ and BaSO4

1 mole of H₂ Ξ 1 mole of BaSO4
0.008928 mole of H₂ Ξ ?? of BaSO4

= (0.008928 mole × 1 mole) / 1 mole
= 0.008928 mole

0.008928 moles of Barium Sulphate was produced.