What mass of aluminum sulfate is required to precipitate all of the Ba?? out of 45.0 mL of 0.548 M barium nitrate solution? 3 Ba(NO3)2 (aq) + Al2(SO4)3 (aq) --> 3 BaSO2 (s) + 2 Al(NO3)3 (aq) ​

2.82 g

Explanation:

Step 1: Write the balanced precipitation reaction

3 Ba(NO₃)₂ (aq) + Al₂(SO₄)₃ (aq) ⇒ 3 BaSO₄(s) + 2 Al(NO₃)₃(aq)

Step 2: Calculate the reacting moles of Ba(NO₃)₂

45.0 mL (0.0450 L) of 0.548 M Ba(NO₃)₂ react.

0.0450 L × 0.548 mol/L = 0.0247 mol

Step 3: Calculate the moles of Al₂(SO₄)₃ that react with 0.0247 moles of Ba(NO₃)₂

The molar ratio of Ba(NO₃)₂ to Al₂(SO₄)₃ is 3:1. The reacting moles of Al₂(SO₄)₃ are 1/3 × 0.0247 mol = 8.23 × 10⁻³ mol

Step 4: Calculate the mass corresponding to 8.23 × 10⁻³ moles of Al₂(SO₄)₃

The molar mass of Al₂(SO₄)₃ is 342.2 g/mol.

8.23 × 10⁻³ mol × 342.2 g/mol = 2.82 g

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