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:

Question

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 0.580g of rhodium (III) sulfate with 0.525g of sodium hydroxide?

Answer 1

This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.
Convert g rhodium sulfate to mols with mols = grams/molar mass
Convert g NaOH to mols the same way.

Using the coefficients in the balanced equation, convert mols of the sulfate to mols of Rh(OH)3.
Do the same with the NaOH to mols Rh(OH)3.
It is likely that you will obtain two different values for mols of the product which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
Convert the smaller value to g Rh(OH)3 by g = mols x molar mass. This is the theoretical yield.