Question

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?

Answer 1

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