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 is the theoretical yield of rhodium(III) hydroxide from the reaction of 0.540 g of rhodium(III) sulfate with 0.209 g of sodium hydroxide?

Question

Quizplus · Accepted Answer

First, we need to determine the limiting reagent to calculate the theoretical yield. To do so, we can compare the number of moles of each reactant:

- moles of Rhodium(III)sulfate = 0.540 g / 203.31 g/mol = 0.002659 mol
- moles of Sodium hydroxide = 0.209 g / 40.00 g/mol = 0.005225 mol

Since the reactants have a 1:6 stoichiometric ratio, we can see that the Rhodium(III)sulfate is the limiting reagent because we would need 0.015954 mol of sodium hydroxide to react with all of the Rhodium(III)sulfate, but we only have 0.005225 mol.

Now that we know the limiting reagent, we can calculate the theoretical yield:

- moles of Rhodium(III)sulfate = 0.002659 mol
- using the stoichiometric ratio, we see that 1 mol Rhodium(III)sulfate produces 2 mol Rhodium(III)hydroxide
- moles of Rhodium(III)hydroxide = 2 x 0.002659 mol = 0.005318 mol
- using the molar mass of Rhodium(III)hydroxide (174.92 g/mol), we can convert moles to grams:
- theoretical yield = 0.005318 mol x 174.92 g/mol = 0.932 g or 0.268 g rounded to three significant figures.

Therefore, the answer is A) 0.268 g.

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