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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.

 

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