Question

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) Of 0.730 g of rhodium(III) hydroxide is produced, what mass of sodium sulfate is also produced?

Answer 1

Answer: The theoretical yield of sodium sulfate is 1.30 grams.

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of rhodium(III) hydroxide = 0.730 g

Molar mass of rhodium(III) hydroxide = 120 g/mol

Putting values in equation 1, we get:

`\text{Moles of rhodium(III) hydroxide}=(0.730g)/(120g/mol)=0.0061mol`

The given chemical equation follows:

`Rh_2(SO_4)_3(aq.)+6NaOH(aq.)\rightarrow 2Rh(OH)_3(s)+3Na_2SO_4(aq.)`

By Stoichiometry of the reaction:

When 2 moles of rhodium(III) hydroxide are produced, then 3 moles of sodium sulfate is also produced

So, when 0.0061 moles of rhodium(III) hydroxide are produced, then =
`(3)/(2)* 0.0061=0.00915mol` of sodium sulfate is also produced

Now, calculating the mass of sodium sulfate from equation 1, we get:

Molar mass of sodium sulfate = 142 g/mol

Moles of sodium sulfate = 0.00915 moles

Putting values in equation 1, we get:

`0.00915mol=\frac{\text{Mass of sodium sulfate}}{142g/mol}\\\\\text{Mass of sodium sulfate}=(0.00915mol* 142g/mol)=1.30g`

Hence, the theoretical yield of sodium sulfate is 1.30 grams.