If a solution containing

55.42

g

55.42 g

of mercury (II) nitrate is allowed to react completely with a solution containing

16.642

g

16.642 g

of sodium sulfate according to the equation below.

Hg

(

NO

3

)

2

(

aq

)

+

Na

2

SO

4

(

aq

)

⟶

2

NaNO

3

(

aq

)

+

H

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

Moles of Hg (NO₃) ₂ = 55.42 / 324.7 = => 0.1707 moles

Moles of Na₂SO₄ = 16.642 / 142.04 = => 0.1172 moles

Limiting reagent is Na₂SO₄ as it controls product formation

Moles of HgSO₄ formed = 0.1172 moles

= 0.1172 x 296.65

= 34.757g