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?
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.
To calculate the percent yield, you need to compare the actual yield (0.330 g) to the theoretical yield.
First, let's determine the theoretical yield of rhodium(III) hydroxide.
1. Calculate the molar mass of Rh2(SO4)3:
Rh2(SO4)3: 2 molar mass of Rh + 3 molar mass of SO4
= (2 * 102.91 g/mol) + (3 * 96.06 g/mol)
= 205.82 g/mol + 288.18 g/mol
= 494.00 g/mol
2. Calculate the number of moles of Rh2(SO4)3:
Moles = Mass / Molar mass
Moles = 0.500 g / 494.00 g/mol
= 0.00101 mol
3. From the balanced chemical equation, we see that the stoichiometric ratio between Rh2(SO4)3 and Rh(OH)3 is 1:2.
So, the number of moles of Rh(OH)3 produced would be twice the number of moles of Rh2(SO4)3:
Moles Rh(OH)3 = 2 * 0.00101 mol
= 0.00202 mol
4. Calculate the theoretical yield of Rh(OH)3:
Mass = Moles * Molar mass
Mass = 0.00202 mol * (3 * 102.91 g/mol + 3 * 1.01 g/mol + 3 * 16.00 g/mol)
= 0.00202 mol * (308.73 g/mol)
= 0.620 g
Now, let's calculate the percent yield:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Percent Yield = (0.330 g / 0.620 g) * 100
= 0.532 * 100
= 53.2%
Therefore, the percent yield of rhodium(III) hydroxide is 53.2%.
To find the percent yield, you need to compare the actual yield (the amount of rhodium(III) hydroxide obtained in the reaction) to the theoretical yield (the maximum amount of rhodium(III) hydroxide that should be produced based on the balanced equation).
First, calculate the theoretical yield of rhodium(III) hydroxide using the molar ratio between rhodium(III) sulfate and rhodium(III) hydroxide from the balanced equation.
1. Convert the given mass of rhodium(III) sulfate to moles:
Molar mass of Rh2(SO4)3 = 2*102.91 g/mol (Rh) + 3*32.07 g/mol (S) + 12*16.00 g/mol (O) = 613.06 g/mol
Number of moles of Rh2(SO4)3 = 0.500 g / 613.06 g/mol
2. Use the balanced equation to determine the molar ratio between Rh2(SO4)3 and Rh(OH)3:
From the equation, it can be observed that 1 mol of Rh2(SO4)3 produces 2 mol of Rh(OH)3.
3. Calculate the theoretical yield of Rh(OH)3:
Theoretical yield of Rh(OH)3 = Number of moles of Rh2(SO4)3 * 2 * molar mass of Rh(OH)3
4. Convert the theoretical yield of Rh(OH)3 from moles to grams.
Now, calculate the percent yield by dividing the actual yield by the theoretical yield, then multiplying by 100 to get the percentage:
Percent yield = (Actual yield / Theoretical yield) * 100
Plug in the given values and solve the equations to find the percent yield.
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