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10 August, 11:06

Sulfuric acid dissolves aluminum metal according to the following reaction: 2Al (s) + 3H2SO4 (aq) →Al2 (SO4) 3 (aq) + 3H2 (g) 2Al (s) + 3H2SO4 (aq) →Al2 (SO4) 3 (aq) + 3H2 (g) Suppose you wanted to dissolve an aluminum block with a mass of 14.6 gg. Part A What minimum mass of H2SO4H2SO4 would you need? Express your answer in grams.

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  1. 10 August, 11:19
    0
    We need 79.6grams of H2SO4

    Explanation:

    Step 1: data given

    Mass of aluminium = 14.6 grams

    Molar mass aluminium = 26.98 g/mol

    Molar mass H2SO4 = 98.08 g/mol

    Step 2: The balanced equation

    2Al (s) + 3H2SO4 (aq) → Al2 (SO4) 3 (aq) + 3H2 (g)

    Step 3: Calculate moles Aluminium

    Moles Aluminium = mass aluminium / molar mass aluminium

    Moles aluminium = 14.6 grams / 26.98 g/mol

    Moles aluminium = 0.541 moles

    Step 4: Calculate moles H2SO4 need

    For 2 moles Al we need 3 moles H2SO4

    For 0.541 moles Al we need 3/2 * 0.541 = 0.8115 moles H2SO4

    Step 5: Calculate mass H2SO4

    Mass H2SO4 = 0.8115 moles * 98.08 g/mol

    Mass H2SO4 = 79.6 grams

    We need 79.6grams of H2SO4
  2. 10 August, 12:14
    0
    76.3 g of H₂SO₄ are needed in this reaction

    Explanation:

    Let's begin with the reaction:

    2Al (s) + 3H₂SO₄ (aq) → Al₂ (SO₄) ₃ (aq) + 3H₂ (g)

    Ratio in the reactants side is 2:3. It means that 2 moles of aluminum need 3 moles of sulfuric acid to react.

    We determine the mass of Al → 14.6 g. 1mol / 26.98 g = 0.519 moles

    Let's propose this rule of three:

    2 moles of Al react with 3 moles of H₂SO₄

    Then 0.519 moles of Al will react with (0.519. 3) / 2 = 0.778 moles of acid.

    We convert the moles to mass, to find the answer:

    0.778 mol. 98 g / 1mol = 76.3 g
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