One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium(III) hydroxide from a solution containing rhodium(III) sulfate according to the following balanced chemical equation:
Rh₂(SO₄) ₃(aq) + 6NaOH(aq) → 2Rh(OH) ₃(s) + 3Na₂SO₄(aq)
If 0.730 g of rhodium(III) hydroxide is produced,what mass of sodium sulfate is also produced?

Question

Quizplus · Accepted Answer

First, calculate the molar mass of Rh(OH)₃: Rh = 102.91 g/mol, O = 16.00 g/mol, H = 1.01 g/mol. So, Rh(OH)₃ = 102.91 + (16.00*3) + (1.01*3) = 165.94 g/mol. Given 0.730 g of Rh(OH)₃, moles of Rh(OH)₃ = 0.730 g / 165.94 g/mol = 0.0044 mol. From the balanced equation, 2 moles of Rh(OH)₃ produce 3 moles of Na₂SO₄. So, 0.0044 mol of Rh(OH)₃ will produce (0.0044 mol * 3/2) = 0.0066 mol of Na₂SO₄. The molar mass of Na₂SO₄ = 2*22.99 + 32.06 + 4*16.00 = 142.04 g/mol. Mass of Na₂SO₄ = 0.0066 mol * 142.04 g/mol = 0.937 g, which rounds to approximately 1.01 g.

One Step in the Isolation of Pure Rhodium Metal (Rh)is