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21 August, 01:09

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

+2
Answers (1)
  1. 21 August, 02:40
    0
    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
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