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 the reaction of 0.500 g of rhodium(III) sulfate with excess sodium hydroxide produces 0.330 g of rhodium(III) hydroxide,what is the percent yield?

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The Answer of 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 the reaction of 0.500 g of rhodium(III)sulfate with excess sodium hydroxide produces 0.330 g of rhodium(III)hydroxide,what is the percent yield? A)423 % B)211 % C)52.9 % D)66 % E)105 %

One Step in the Isolation of Pure Rhodium Metal (Rh)is