HURRY Show step-by-step:
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:

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
0.540-g of rhodium (III) sulfate with 0.209-g of sodium hydroxide?

The answer is 0.268-g but HOW? Please show how to solve this.

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