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Question
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)
If 2.40g of rhodium(III) sulfate reacts with excess sodium hydroxide, what mass of rhodium(III) hydroxide may be produced?
A) 1.50 g
B) 4.80 g
C) 2.40 g
D) 0.374 g
E) 2.99 g
Answer
This answer is hidden. It contains 1 characters.
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