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 0.580g of rhodium (III) sulfate with 0.525g of sodium hydroxide?
This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.
Convert g rhodium sulfate to mols with mols = grams/molar mass
Convert g NaOH to mols the same way.
Using the coefficients in the balanced equation, convert mols of the sulfate to mols of Rh(OH)3.
Do the same with the NaOH to mols Rh(OH)3.
It is likely that you will obtain two different values for mols of the product which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
Convert the smaller value to g Rh(OH)3 by g = mols x molar mass. This is the theoretical yield.
To determine the theoretical yield of rhodium (III) hydroxide, we need to find the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed and limits the amount of product formed.
1. Calculate the number of moles for each reactant:
- Moles of Rh2(SO4)3 = (0.580 g) / (molar mass of Rh2(SO4)3)
- Moles of NaOH = (0.525 g) / (molar mass of NaOH)
2. Look at the balanced chemical equation to determine the stoichiometry of the reaction. From the equation, we can see that the molar ratio between Rh2(SO4)3 and Rh(OH)3 is 2:2 (or 1:1).
3. Compare the mole ratios calculated in step 1 with the stoichiometry in step 2. The reactant with the smaller mole ratio is the limiting reagent.
4. Once you determine the limiting reagent, you can calculate the theoretical yield by using the stoichiometric ratio between the limiting reagent and the desired product.
Let's assume the molar mass of Rh2(SO4)3 is x and the molar mass of NaOH is y.
- Moles of Rh2(SO4)3 = (0.580 g) / (x g/mol)
- Moles of NaOH = (0.525 g) / (y g/mol)
- Multiply the moles of Rh2(SO4)3 by the stoichiometric ratio (2 mol Rh(OH)3 / 2 mol Rh2(SO4)3) to find the moles of Rh(OH)3.
The theoretical yield in grams can be calculated by:
Theoretical yield (g) = (moles of Rh(OH)3) x (molar mass of Rh(OH)3)
Please provide the molar masses of Rh2(SO4)3 and NaOH so I can complete the calculation.
To determine the theoretical yield of rhodium (III) hydroxide (Rh(OH)3), we need to calculate the amount of reagent that limits the reaction, also known as the limiting reagent.
Step 1: Convert the given masses of the reactants to moles using their respective molar masses.
- Mass of Rhodium (III) sulfate (Rh2(SO4)3) = 0.580g
- Molar mass of Rh2(SO4)3 = (2 * Atomic mass of Rh) + (3 * Atomic mass of S) + (12 * Atomic mass of O) = (2 * 102.91) + (3 * 32.07) + (12 * 16.00) = 469.1 g/mol
- Moles of Rh2(SO4)3 = mass / molar mass = 0.580g / 469.1 g/mol
- Mass of sodium hydroxide (NaOH) = 0.525g
- Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 39.99 g/mol
- Moles of NaOH = mass / molar mass = 0.525g / 39.99 g/mol
Step 2: Determine the stoichiometric ratio of the reactants.
According to the balanced chemical equation:
Rh2(SO4)3(aq) + 6NaOH(aq) --> 2Rh(OH)3(s) + 3Na2SO4(aq)
The ratio between Rh2(SO4)3 and Rh(OH)3 is 1:2. This means that for every 1 mole of Rh2(SO4)3, we should get 2 moles of Rh(OH)3.
Step 3: Calculate the mole ratio between the reactants and identify the limiting reagent.
- Moles of Rh2(SO4)3 = 0.580g / 469.1 g/mol = 0.0012 mol (rounded to four decimal places)
- Moles of NaOH = 0.525g / 39.99 g/mol = 0.0131 mol (rounded to four decimal places)
Since the mole ratio between Rh2(SO4)3 and Rh(OH)3 is 1:2, we need 1/2 the number of moles of Rh2(SO4)3 to form the same number of moles of Rh(OH)3. In this case, we have fewer moles of Rh2(SO4)3 than NaOH. Therefore, Rh2(SO4)3 is the limiting reagent.
Step 4: Calculate the theoretical yield of Rh(OH)3.
Since 1 mole of Rh(OH)3 corresponds to 0.0012 moles of Rh2(SO4)3, we can calculate the theoretical yield as follows:
Theoretical yield = (moles of limiting reagent) × (molar mass of Rh(OH)3)
= 0.0012 mol × (molar mass of Rh(OH)3)
The molar mass of Rh(OH)3 is calculated as follows:
- Molar mass of Rh = 102.91 g/mol
- Molar mass of O = 16.00 g/mol
- Molar mass of H = 1.01 g/mol
- Molar mass of Rh(OH)3 = (2 * 102.91) + (3 * 1.01) + (3 * 16.00) = 249.12 g/mol
Theoretical yield = 0.0012 mol × 249.12 g/mol
Therefore, the theoretical yield of Rh(OH)3 is calculated to be (0.0012 mol) × (249.12 g/mol) = 0.299 g (rounded to three decimal places).