Question

A solution containing 87.32g of dissolved mercury (III) nitrate is mixed with a solution containing 34.62g of dissolved sodium sulfate. A chemical reaction occurs, and a precipitate is produced. How many grams of solid are produced from the reaction and how many grams of excess reactant remain after the reaction is complete?

Answer 1

Mercury (III) nitrate will not exist,but mercury (II) nitrate exist.

Hg(NO₃)₂ (aq)+ Na₂SO₄ (aq) ---> HgSO₄ (s)+ 2NaNO₃ (aq)

Molar mass of Hg(NO₃)₂ = 200.6+ [2x(14+(3x16))] = 324.6 g/mol

Molar mass of Na₂SO₄ = (2x23) + 32+(4x16)

                                  = 142 g/mol

MOlar mass of HgSO4 is = 200.6 +32+(4x16) = 296.6 g/mol

According to the balanced equation ,

1 mole of Hg(NO₃)₂ reacts with 1 mole of Na₂SO₄

324.6 g of Hg(NO₃)₂ reacts with 142 g of Na₂SO₄

M g of Hg(NO₃)₂ reacts with 34.62 g of Na₂SO₄

M = (324.6x34.62) / 142

   = 79.14 g of Hg(NO₃)₂

So 87.32 - 79.14 = 8.18 g of Hg(NO₃)₂ left unreacted so Hg(NO₃)₂ is the excess reactant.

Since all the mass of Na₂SO₄ completly reacted it is the limiting reactant.

From the balanced reaction,

1 mole of Na₂SO₄ produces 1 mole of HgSO₄

                           OR

142 g of Na₂SO₄ produces 296.6 g of HgSO₄

34.62 g of Na₂SO₄ produces N g of HgSO₄

N = (34.62x296.6) / 142

   = 72.31 g of HgSO₄

Therefore the mass of solid precipitated is 72.31 g