Answer to Question #87590 in General Chemistry for Fasika

Question

If a solution containing 57.620 g of mercury(II) perchlorate is allowed to react completely with a solution containing 15.488 g 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?

Answer 1

 X mol       0.059 mol     Y mol

Hg(ClO₄)₂ + Na₂Cr₂O₇ = HgCr₂O₇↓ + 2NaClO₄

 1 mol           1 mol          1 mol

n(Hg(ClO₄)₂) = m(Hg(ClO₄)₂)/M(Hg(ClO₄)₂) = 57.620 g / 400 g/mol = 0.144 mol – excess.

n(Na₂Cr₂O₇) = m(Na₂Cr₂O₇)/M(Na₂Cr₂O₇) = 15.488 g / 262 g/mol = 0.059 mol – deficiency.

n(HgCr₂O₇) = Y = 0.059 mol.

m(HgCr₂O₇) = M(HgCr₂O₇)×n(HgCr₂O₇) = 417 g/mol × 0.059 mol = 24.603 g.

n_exc(Hg(ClO₄)₂) = n(Hg(ClO₄)₂) – X = 0.144 mol – 0.059 mol = 0.085 mol.

m_exc(Hg(ClO₄)₂) = M(Hg(ClO₄)₂)×n_exc(Hg(ClO₄)₂) = 400 g/mol × 0.085 mol = 34 g.