Final answer:
To find the mass of sodium sulfate produced from 0.510 g of rhodium(III) hydroxide, calculate the moles of rhodium(III) hydroxide, then use the stoichiometry of the chemical equation to find the moles of sodium sulfate. Finally, convert the moles of sodium sulfate to grams, which gives approximately 0.518 g of sodium sulfate.
Step-by-step explanation:
To calculate the mass of sodium sulfate produced when 0.510 g of rhodium(III) hydroxide is produced, we need to use a stoichiometric approach based on the balanced chemical equation. The equation given is:
RH₂(SO₄)₃(aq) + 6NaOH(aq) → 2Rh(OH)₃(s) + 3Na₂SO₄(aq)
From this equation, we can see that each mole of rhodium(III) sulfate produces 1.5 moles of sodium sulfate.
First, we have to calculate the moles of rhodium(III) hydroxide produced:
Rh(OH)₃ has a molar mass of about 209.88 g/mol.
0.510 g Rh(OH)₃ × (1 mol / 209.88 g) = 0.00243 mol Rh(OH)₃
Using the molar ratio from the balanced equation:
0.00243 mol Rh(OH)₃ × (1.5 mol Na₂SO₄ / 1 mol Rh(OH)₃) = 0.00365 mol Na₂SO₄
Now, we calculate the mass of sodium sulfate using its molar mass (Na₂SO₄ has a molar mass of about 142.04 g/mol):
0.00365 mol Na₂SO₄ × (142.04 g / 1 mol) = 0.518 g Na₂SO₄
Therefore, 0.510 g of rhodium(III) hydroxide will produce approximately 0.518 g of sodium sulfate.