Rh2(SO4)3 + 6NaOH = 2Rh(OH)3 + 3Na2SO4
So you expect to get twice as many moles of Rh(OH)3 as you used of Rh2(SO4)3
Rh2(SO4)3 has a mol wt of 494, so you have 0.5/494 = 0.001012 moles
Now see how many moles of output you get, and
divide what you got by what you expected.
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: Rh 2 (SO 4 ) 3 (aq)+6NaOH(aq) 2Rh(OH) 3 (s)+3Na 2 SO 4 (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?
2 answers
molar masses:
Rh = 103 grams/mol
S = 32 g/mol
O = 16 g/mol
H atom = 1 g/mol
so
Rh 2 (SO 4 ) 3 = 2*103 + 3[32+ 4(16)] = 206+3[96] = 494 g/mol
Rh(OH) 3 = 103 +3*17 = 154 so TWO mols is 308
so for every 494 grams you get 308 grams in perfect world
so for .5 grams you should get .3117 grams
.330 is better than 100% yield, error somewhere
Rh = 103 grams/mol
S = 32 g/mol
O = 16 g/mol
H atom = 1 g/mol
so
Rh 2 (SO 4 ) 3 = 2*103 + 3[32+ 4(16)] = 206+3[96] = 494 g/mol
Rh(OH) 3 = 103 +3*17 = 154 so TWO mols is 308
so for every 494 grams you get 308 grams in perfect world
so for .5 grams you should get .3117 grams
.330 is better than 100% yield, error somewhere
2 answers