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

Answer 1

Molar mass of Rh2(SO4)3 = 493.9988 g/mol
Molar mass of NaOH=39.997 g/mol
Molar mass of Rh(OH)3=153.92763g/mol

Find the limiting reagent, which is the reactant with the lowest number of moles.

0.580 g of rhodium(III) sulfate *(1 mole/493.9988 g)=moles of Rh2(SO4)3

0.525 g of sodium hydroxide *(1 mole/39.997 g)= moles of NaOH

Eyeballing it, I think it is Rh2(SO4)3

moles of Rh2(SO4)3*(2 moles of Rh(OH)3/1 mole of Rh2(SO4)3)= moles of Rh(OH)3

moles of Rh(OH)3*(153.92763g/mole)= theoretical yield of rhodium(III) hydroxide

Answer should have 3 significant figures

Answer 2

I got .369 g Rh(OH)3. RIGHT?