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?

The theoretical yield of sodium sulfate is 1.30 grams.

Explanation:

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

     .....(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:

The given chemical equation follows:

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 = 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:

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

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