Question

A laboratory technician combines 26.0 mL of 0.379 M cobalt(II) bromide with 32.1 mL 0.376 M potassium hydroxide. How many grams of cobalt(II) hydroxide can precipitate?

__g

Answer 1

CoBr2(aq) + 2KOH(aq) <-------> Co(OH)2(s) + 2KBr(aq)

Stoichiometrically, 2mol of KOH react with 1mole of CoBr2

No of mole of CoBr2 = (0.379mol/1000ml)×26.0ml=0.009854

No of mole of KOH = (0.376mol/1000ml)×32.1ml= 0.01207

0.009854 mol require 2×0.009854=0.019708mol of KOH but available mole of KOH is 0.01207

KOH is limiting

Stoichiometrically, 2mol of KOH give 1mol of Co(OH)2

Therefore,

0.01207 mol of KOH should give 0.01207/2 =0.00635mol of Co(OH)2

Molar mass of Co(OH)2 = 92.948g/mol

Mass of Co(OH)2 produced = 92.948g/mol × 0.00635mol

= 0.5902g