Question

If a solution containing 17.30 g of mercury(II) nitrate is allowed to react completely with a solution containing 5.102 g of sodium sulfate, how many grams of solid precipitate will be formed? BHow many grams of the reactant in excess will remain after the reaction?

Answer 1

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

Molar mass of Hg(NO₃)₂ is = At.mass of Hg + (2xAt.mass of N ) + (6xAt.mass of O)

                                         = 200.6 + (2x14) + (6x16)

                                         = 324.6 g/mol

Molar mass of Na₂SO₄ = (2xAt.mass of Na ) + At.mass of S + (4xAt.mass of O)

                                  = (2x23) + 32 + ( 4x16)

                                  = 142 g/mol

Molar mass of HgSO₄ = At.mass of Hg + At.mass of S + (4xAt.mass of O)

                                = 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

                              OR

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

M g of Hg(NO₃)₂ reacts with 5.102 g of Na₂SO₄      

M = (324.6x5.102) / 142

    = 11.66 g

So 17.30-11.66 = 5.64 g of mercury(II) nitrate will remian unreacted which is the excess reactant.

Therefore the mass of excess reactant(mercury(II) nitrate) remained is 5.64 g

According to the balanced equation,

1 mole of Na₂SO₄ produces 1 mole of HgSO₄  

                             OR

142 g of Na₂SO₄ produces 296.6 g of HgSO₄  

5.102 g of Na₂SO₄ produces N g of HgSO₄  

N = (5.102x296.6)/142

   = 10.66 g of HgSO₄