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.

they both have a y in them

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.