If a solution containing 33.82 g of mercury(II) acetate is allowed to react completely with a solution containing 9.718 g of sodium sulfate, how many grams of solid precipitate will be formed?

How many grams of the reactant in excess will remain after the reaction?

To find the number of grams of solid precipitate formed and the grams of the excess reactant remaining after the reaction, we need to first determine the limiting reactant.

1. Convert the given masses of mercury(II) acetate and sodium sulfate to moles.
- Mercury(II) acetate: Mercury(II) acetate has the chemical formula Hg(C2H3O2)2, with a molar mass of 318.65 g/mol. Therefore, the moles of mercury(II) acetate can be calculated as:
moles of Hg(C2H3O2)2 = mass / molar mass = 33.82 g / 318.65 g/mol
- Sodium sulfate: Sodium sulfate has the chemical formula Na2SO4, with a molar mass of 142.04 g/mol. Therefore, the moles of sodium sulfate can be calculated as:
moles of Na2SO4 = mass / molar mass = 9.718 g / 142.04 g/mol

2. Determine the stoichiometric ratio between the reactants. Based on the balanced equation for the reaction between mercury(II) acetate and sodium sulfate:
Hg(C2H3O2)2 + Na2SO4 -> HgSO4 + 2NaC2H3O2
we can see that one mole of mercury(II) acetate reacts with one mole of sodium sulfate to produce one mole of mercury(II) sulfate.

3. Compare the moles of each reactant to identify the limiting reactant. The reactant that produces the least amount of product is the limiting reactant.
moles of mercury(II) acetate = 33.82 g / 318.65 g/mol ≈ 0.106 mol
moles of sodium sulfate = 9.718 g / 142.04 g/mol ≈ 0.068 mol

Since sodium sulfate produces fewer moles of product, it is the limiting reactant.

4. Calculate the moles of solid precipitate formed. From the reaction stoichiometry, we know that 1 mole of sodium sulfate reacts with 1 mole of mercury(II) sulfate. Therefore, the moles of solid precipitate formed will be equal to the moles of sodium sulfate used in the reaction, which is approximately 0.068 mol.

5. Convert the moles of solid precipitate to grams. The molar mass of mercury(II) sulfate (HgSO4) is 296.65 g/mol.
mass of solid precipitate = moles of HgSO4 * molar mass of HgSO4 = 0.068 mol * 296.65 g/mol

So, the grams of solid precipitate formed will be approximately 20.170 g.

6. Calculate the grams of the excess reactant remaining. Since sodium sulfate is the limiting reactant, all of it will be consumed in the reaction. Therefore, none of the sodium sulfate will remain after the reaction.

To summarize:
- Grams of solid precipitate formed: approximately 20.170 g.
- Grams of sodium sulfate remaining: 0 g.