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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?
Oct 11th, 2015

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Best Answer:Rh2(SO4)3 + 6 NaOH → 2 Rh(OH)3 + 3 Na2SO4 

(0.760 g Rh2(SO4)3) / (494.0007 g/mol) = 0.00153845936 mol Rh2(SO4)3 
(0.473 g NaOH) / (39.9971 g/mol) = 0.0118258574 mol NaOH 

0.00153845936 mol Rh2(SO4)3 would react completely with 6 x 0.00153845936 = 0.00923075616 mol NaOH, but there is more NaOH than that present, so NaOH is in excess and Rh2(SO4)3 is the limiting reactant. 

(0.00153845936 mol Rh2(SO4)3 ) x (2/1) x (153.9276 g/mol) = 0.474 g Rh(OH)3

Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 11th, 2015

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