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)
What mass of sodium hydroxide is required to precipitate 68.0 g of rhodium(III) hydroxide from a solution containing excess rhodium(III) sulfate?

Question

Quizplus · Accepted Answer

According to the balanced chemical equation, 1 mole of Rh(OH)3 is produced for every 6 moles of NaOH used.
We can first convert the given mass of Rh(OH)3 to moles:
68.0 g Rh(OH)3 × (1 mol Rh(OH)3/159.91 g Rh(OH)3) = 0.425 mol Rh(OH)3

Since there is excess Rh2(SO4)3, we can assume that all of the Rh2(SO4)3 is converted to Rh(OH)3, and thus we can use stoichiometry to find the amount of NaOH required.

0.425 mol Rh(OH)3 × (6 mol NaOH/1 mol Rh(OH)3) × (40.00 g NaOH/1 mol NaOH) = 102 g NaOH (to 3 significant figures)

Therefore, the mass of NaOH required to precipitate 68.0 g of Rh(OH)3 is 102 g, which corresponds to choice (B).

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