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?

The mass of the solid precipitate (HgSO₄) is 27.529 g

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

The mass of the reactant that remained (Hg(NO₃)₂) after the reaction is 16.133 g

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

the correct answer would be d. hope this !

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