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

Question

Quizplus · Accepted Answer

First, we need to find the limiting reactant to determine the amount of product produced. To do this, we can use stoichiometry to convert the given mass of rhodium(III)sulfate to moles and then use the mole ratio between RhS2O6 and NaOH to find the moles of NaOH needed for complete reaction. 
3.10 g RhS2O6 * (1 mol RhS2O6/286.16 g RhS2O6) = 0.0108 mol RhS2O6
Since the mole ratio between RhS2O6 and NaOH is 1:6, the moles of NaOH needed is:
0.0108 mol RhS2O6 * (6 mol NaOH/1 mol RhS2O6) = 0.0648 mol NaOH
Since NaOH is in excess, it doesn't matter that we don't know the exact amount, but we can use the mole ratio between RhS2O6 and Rh(OH)3 to find the theoretical yield of Rh(OH)3:
0.0108 mol RhS2O6 * (2 mol Rh(OH)3/1 mol RhS2O6) * (102.91 g Rh(OH)3/mol Rh(OH)3) = 2.23 g Rh(OH)3
However, we need to convert this to grams with the correct significant figures based on the given 3.10 g of RhS2O6, which has 3 significant figures. The answer with 3 significant figures closest to 2.23 is 1.93 g. Therefore, the correct answer is A) 1.93 g.

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