Question

How many grams are needed of ammonium hydroxide if it is reacted with bismuth(III) sulfate to give you 16 grams of bismuth hydroxide, the other product is ammonium sulfate. Also how many grams of bismuth(III) sulfate are needed?

Answer 1

The reaction can be represented as following:

6NH₄OH + Bi₂(SO₄)₃ --> 2Bi(OH)₃ + 3(NH₄)₂SO₄

From the reaction:

m(NH₄OH) / [6 × Mr(NH₄OH)] = m(Bi(OH)₃) / [2 × Mr(Bi(OH)₃)] = m(Bi₂(SO₄)₃) / Mr(Bi₂(SO₄)₃)

From here:

m(NH₄OH) = [m(Bi(OH)₃) × 6 × Mr(NH₄OH)] / [2 × Mr(Bi(OH)₃)] = [16g × 6 × 35] / [2 × 260 g/mol] = 6.46 g

m(Bi₂(SO₄)₃) = [m(Bi(OH)₃) × Mr(Bi₂(SO₄)₃)] / [2 × Mr(Bi(OH)₃)] = [16 g × 706 g/mol] / [2 × 260 g/mol] = 21.72 g

Answer: 6.46 g of NH₄OH and 21.72 g of Bi₂(SO₄)₃.