Problem- 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 0.590 g of rhodium(III) sulfate with 0.266 g of sodium hydroxide?
I want you to clarify the chemistry in this above problem.