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?
(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
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