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

Question

Quizplus · Accepted Answer

To solve this, first, calculate the molar mass of rhodium(III) sulfate, $\text{Rh}_2(\text{SO}_4)_3$, and rhodium(III) hydroxide, $\text{Rh(OH)}_3$. The molar mass of $\text{Rh}_2(\text{SO}_4)_3$ is approximately 482.96 g/mol (Rh = 102.91 g/mol, S = 32.07 g/mol, O = 16.00 g/mol), and the molar mass of $\text{Rh(OH)}_3$ is approximately 165.93 g/mol. According to the balanced equation, 2 moles of $\text{Rh(OH)}_3$ are produced per mole of $\text{Rh}_2(\text{SO}_4)_3$. $$3.20 \text{ g} \times \frac{1 \text{ mol}}{482.96 \text{ g}} \times 2 \text{ mol Rh(OH)}_3 \times 165.93 \text{ g/mol} \approx 1.99 \text{ g}$$This calculation shows that 1.99 g of rhodium(III) hydroxide may be produced.

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