Question

If a solution containing 33.53 g of mercury(II) acetate is allowed to react completely with a solution containing 9.718 g of sodium dichromate, how many grams of solid precipitate will form

Answer 1

14.58 g

Step-by-step explanation:

Hg(C2H3O2)2(aq) + NaCr2O7(aq) ------> 2NaC2H3O2(aq) + HgCr2O7(s)

Number of moles of mercury II acetate = 33.53g/318.7 g/mol = 0.11 moles

Number of moles of sodium dichromate = 9.178 g/261.97 g/mol = 0.035 moles

Since the molar ratio is 1:1, sodium dichromate is the limiting reactant.

Molar mass of mercuric chromate = 416.58 g/mol

Mass of 416.58 g/mol produced = 0.035 moles * 416.58 g/mol = 14.58 g