Solid Precipitate/ Excess Reactant Chem Question

Question

If a solution containing 27.24 g of mercury(II) acetate is allowed to react completely with a solution containing 8.564 g of sodium dichromate, how many grams of solid precipitate will form?

Answer 1

First, write the correctly balanced equation for the reaction taking place:

Hg(CH₃COO)₂(aq) + Na₂Cr₂O₇(aq) ==> 2NaCH₃COO(aq) + HgCr₂O₇(s)

Information needed:

molar mass Na₂Cr₂O₇ = 261.97 g/mol

molar mass Hg(CH₃COO)₂ = 318.68 g/mol

molar mass HgCr₂O₇ = 416.58 g/mol

1). Find the limiting reactant. One way to do this is to divide the moles of each reactant by the corresponding coefficient in the balanced equation, and which ever value is less represents the limiting reactant:

For Hg(CH₃COO)₂: 27.24 g x 1 mol / 318.68 g = 0.08548 moles

For Na₂Cr₂O₇: 8.564 g x 1 mol / 261.97 g = 0.03269 moles

Since both reactants have a coefficient of 1 in the balanced equation, Na₂Cr₂O₇ is the limiting reactant.

2). Use the moles of the limiting reactant and stoichiometry to find the moles of precipitate, HgCr₂O₇(s):

0.03269 mols Na₂Cr₂O₇ x 1 mol HgCr₂O₇ / mol Na₂Cr₂O₇ = 0.03269 mols HgCr₂O₇

3). Use molar mass of HgCr₂O₇ to convert moles to grams:

0.03269 mols HgCr₂O₇ x 416.58 g / mol = 13.62 g HgCr₂O₇