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 6.20 g of rhodium(III) sulfate reacts with excess sodium hydroxide,what mass of rhodium(III) hydroxide may be produced?

Question

Quizplus · Accepted Answer

The balanced chemical equation shows that 1 mole of Rh2(SO4)3 reacts with 6 moles of NaOH to produce 2 moles of Rh(OH)3. The molar mass of Rh2(SO4)3 is 458.24 g/mol, and the molar mass of Rh(OH)3 is 103.02 g/mol. Therefore, 6.20 g of Rh2(SO4)3 will react to produce 6.20 g / 458.24 g/mol * 2 mol Rh(OH)3 / 1 mol Rh2(SO4)3 * 103.02 g/mol Rh(OH)3 = 3.86 g of Rh(OH)3.

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