Answer:
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
