Question

If a solution containing 48.32g of mercury(II) acetate is allowed to react completely with a solution containing 15.488g of sodium dichromate, how many grams of solid precipitate will be formed?

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

Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (0) for the number of moles.

C2H3O–2

Na+

Cr2O2−7

Answer 1

How many grams of solid precipitate will be formed? : 24.63 g

How many grams of the reactant in excess will remain after the reaction? : 29.48 g

C₂H₃O₂^- = 0.303 mol

Na⁺ = 0.118 mol

Cr₂O₇^2- = 0

Hg²⁺ = 0.0925 mol

Explanation

The balanced reaction equation is

Hg(C₂H₃O₂)₂ (aq) + Na₂Cr₂O₇ (aq) HgCr₂O₇ (s) + 2 NaC₂H₃O₂ (aq)

For the given amounts of reactants, sodium dichromate Na₂Cr₂O₇ is the limiting reactant.