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 0.580 g of rhodium(III) sulfate with 0.525 g of sodium hydroxide?
To find the theoretical yield of rhodium(III) hydroxide, we need to calculate the limiting reactant and use it to determine the maximum amount of product that can be formed.
Step 1: Determine the molar masses of the reactants
- Molar mass of Rh2(SO4)3: (2 x atomic mass of Rh) + (3 x atomic mass of S) + (12 x atomic mass of O) + (4 x atomic mass of O)
- Molar mass of NaOH: atomic mass of Na + atomic mass of O + atomic mass of H
Step 2: Convert the masses of the reactants to moles
- Moles of Rh2(SO4)3 = mass of Rh2(SO4)3 / molar mass of Rh2(SO4)3
- Moles of NaOH = mass of NaOH / molar mass of NaOH
Step 3: Use the balanced chemical equation to determine the mole ratio between the reactants and the product
According to the balanced chemical equation:
- 1 mole of Rh2(SO4)3 produces 2 moles of Rh(OH)3
- 6 moles of NaOH produces 2 moles of Rh(OH)3
Step 4: Determine the limiting reactant
The limiting reactant is the one that produces the least amount of product. To find the limiting reactant, compare the mole ratios from step 3 and see which reactant has a smaller ratio. The reactant with the smaller ratio is the limiting reactant.
Step 5: Calculate the theoretical yield of Rh(OH)3
- If Rh2(SO4)3 is the limiting reactant:
- Use the mole ratio from step 3 to calculate the moles of Rh(OH)3 produced
- Convert moles of Rh(OH)3 to grams using the molar mass of Rh(OH)3
- If NaOH is the limiting reactant:
- Use the mole ratio from step 3 to calculate the moles of Rh(OH)3 produced
- Convert moles of Rh(OH)3 to grams using the molar mass of Rh(OH)3
The theoretical yield of Rh(OH)3 is the smaller value obtained from the two calculations above.
To find the theoretical yield of rhodium(III) hydroxide, we need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
To determine the limiting reactant, we compare the amount of rhodium(III) sulfate (Rh2(SO4)3) and sodium hydroxide (NaOH) used in the reaction.
1) Convert the masses of the reactants to moles:
0.580 g Rh2(SO4)3 x (1 mol Rh2(SO4)3 / molar mass of Rh2(SO4)3)
0.525 g NaOH x (1 mol NaOH / molar mass of NaOH)
2) Use the stoichiometric coefficients from the balanced chemical equation to determine the mole ratio between the reactants:
Rh2(SO4)3 : NaOH = 1 : 6
3) Compare the moles of the reactants using the mole ratio calculated in step 2 to determine the limiting reactant.
4) Once the limiting reactant is known, we can use the stoichiometry from the balanced equation to calculate the theoretical yield of rhodium(III) hydroxide.
Finally, substitute the limiting reactant's moles into the stoichiometric relationship to calculate the theoretical yield of rhodium(III) hydroxide:
Theoretical yield of Rh(OH)3 = (moles of limiting reactant) x (molar ratio of Rh(OH)3 / limiting reactant) x (molar mass of Rh(OH)3)
Note: The molar mass of Rh(OH)3 is required to calculate the answer, so it should already be known.
Molar mass of Rh2(SO4)3 = 493.9988 g/mol
Molar mass of NaOH=39.997 g/mol
Molar mass of Rh(OH)3=153.92763g/mol
Find the limiting reagent, which is the reactant with the lowest number of moles.
0.580 g of rhodium(III) sulfate *(1 mole/493.9988 g)=moles of Rh2(SO4)3
0.525 g of sodium hydroxide *(1 mole/39.997 g)= moles of NaOH
Eyeballing it, I think it is Rh2(SO4)3
moles of Rh2(SO4)3*(2 moles of Rh(OH)3/1 mole of Rh2(SO4)3)= moles of Rh(OH)3
moles of Rh(OH)3*(153.92763g/mole)= theoretical yield of rhodium(III) hydroxide
Answer should have 3 significant figures