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)

Question

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) ​

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Delwyn 4 years 2021-08-04T22:33:10+00:00 1 Answers 10 views 0

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  1. Nem
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    2021-08-04T22:34:49+00:00
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    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

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