Question

Based on 6.8 grams of solid sodium hydroxide reacts with excess aqueous sulfurous acid (H2SO3) react to form solid sodium sulfite and liquid water.

Find moles of H20 expected to be produced.

Find the mass of H20 expected to be produced.

Answer 1

Balanced equation for the given reaction is,

H₂SO₃(aq) + 2NaOH(s) = Na₂SO₃(aq) + 2H₂O(l)

We are given that excess of sulfurous acid is present thus NaOH will act as a limiting reactant.

According to the coeffecient ratio rule we know that in given reaction, NaOH and H₂O react in a ratio of 1 : 1 i.e. for every 1 mole of NaOH 1 mole of H₂O is produced.

No. of moles of NaOH in solution = Mass of NaOH given / Molar mass of NaOH

Mass of NaOH given = 6.8 g

Molar mass of NaOH = 40 g/mol

No. of moles of NaOH in solution = (6.8 g ) / (40 g/mol)

No. of moles of NaOH in solution = 0.17 mol

As, NaOH and H₂O react in a ratio of 1 : 1_,

No. of moles of H₂O expected to be produced = 0.17 mol

Molar mass of H₂O = 18 g/mol

Mass of H₂O in solution = (No. of moles of H₂O in solution) * (Molar mass of H₂O)

Mass of H₂O in solution = (0.17 mol) * (18 g/mol)

Mass of H₂O expected to be produced = 3.06 g