If a solution containing 46.26 g of mercury (II) nitrate is allowed to react completely with a solution containing 13.180 g of sodium sulfate according to the equation below. Hg (NO3) 2 (aq) + Na2SO4 (aq) ⟶2NaNO3 (aq) + HgSO4 (s) Hg (NO 3) 2 (aq) + Na 2 SO 4 (aq) ⟶ 2 NaNO 3 (aq) + HgSO 4 (s) How many grams of solid precipitate will be formed? How many grams of the reactant in excess will remain after the reaction?

1 (a). Answer:

The mass of the solid precipitate (HgSO₄) is 27.529 g

Solution:

The equation for the reaction;

Hg (NO₃) ₂ (aq) + Na₂SO₄ (aq) ⟶2NaNO₃ (aq) + HgSO₄ (s)

Step 1: Moles of mercury (II) nitrate and moles of sodium sulfate

Number of moles = Mass / Molar mass

Moles of Hg (NO₃) ₂ = 46.26 g : 324.60 g/mol

= 0.1425 moles

Moles of Na₂SO₄ = 13.180 g : 142.04 g/mol

= 0.0928 moles

According to the equation 1 mole of Hg (NO₃) ₂ reacts with 1 mole Na₂SO₄ therefore, 0.0928 moles of Na₂SO₄ will react with 0.0928 moles of Hg (NO₃) ₂

.

Step 2: Moles of the solid precipitate (HgSO₄)

The mole ratio of Na₂SO₄ to HgSO₄ (s) is 1:1

Therefore; the moles of HgSO₄ is 0.0928 moles

Step 3: Mass of the solid precipitate formed (HgSO₄)

Mass = Number of moles * Molar mass

= 0.0928 moles * 296.65 g/mol

= 27.529 g

1 (b). Answer:

The mass of the reactant that remained (Hg (NO₃) ₂) after the reaction is 16.133 g

Solution According to the equation 1 mole of Hg (NO₃) ₂ reacts with 1 mole Na₂SO₄ therefore, 0.0928 moles of Na₂SO₄ will react with 0.0928 moles of Hg (NO₃) ₂. Thus, Hg (NO₃) ₂ is the excess reactant.

Mass of the reactant in excess after the reaction.

Hg (NO₃) ₂ is the reactant that was in excess.

0.0928 moles of Hg (NO₃) ₂ reacted with Na₂SO₄.

Therefore;

Remaining number of moles of Hg (NO₃) ₂ = 0.1425 moles - 0.0928 moles

= 0.0497 moles

Mass of Hg (NO₃) ₂ that remained after the reaction

= 0.0497 moles * 324.60 g/mol

= 16.133 g