What mass of Na2SO4 will be formed by addition of 17.0 g of NaHCO3 in aqueous solution to an aqueous solution containing 0.400 mol of H2SO4?

Mass of Na₂SO₄= 14.346 g

Given

17 g NaHCO₃

0.4 mol H₂SO₄

Required

mass of Na₂SO₄

Solution

Reaction

H₂SO₄ (aq) + 2 NaHCO₃ (aq) ⇒ Na₂SO₄(aq) + 2 CO₂(g) + 2 H₂O (l)

mol NaHCO₃ :

= mass : MW

= 17 g : 84 g/mol

= 0.202

Find limiting reactant = mol : coefficient

H₂SO₄ = 0.4 :  1 = 0.4

NaHCO₃ = 0.202 : 2 = 0.101

NaHCO₃ as a limiting reactant (smaller ratio)

Mol Na₂SO₄ based on mol NaHCO₃

From the equation, mol ratio NaHCO₃ : Na₂SO₄ = 2 : 1, so mol Na₂SO₄ :

= 1/2 x mol NaHCO₃

= 1/2 x 0.202

= 0.101

Mass of Na₂SO₄ :

= mol x MW

= 0.101 x 142,04 g/mol

= 14.346 g

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