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20 May, 23:15

Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to the reaction BaO 2 (s) + H 2 SO 4 (aq) ⟶ BaSO 4 (s) + H 2 O 2 (aq) BaO2 (s) + H2SO4 (aq) ⟶BaSO4 (s) + H2O2 (aq) How many milliliters of 3.00 M H 2 SO 4 (aq) 3.00 M H2SO4 (aq) are needed to react completely with 92.5 g BaO 2 (s) ? 92.5 g BaO2 (s) ?

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  1. 20 May, 23:28
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    There is 182 mL of 3.00 M H2SO4 needed

    Explanation:

    Step 1: Data given

    Mass of BaO2 = 92.5 grams

    Molarity of 3.00 M H2SO4

    Step 2: The balanced equation

    BaO2 (s) + H2SO4 (aq) ⟶ BaSO4 (s) + H2O2 (aq)

    For 1 mole of BaO2 we need 1 mole of H2SO4 to produce 1 mole of BaSO4 and 1 mole H2SO2

    Step 3: Calculate moles of BaO2

    Moles BaO2 = mass BaO2 / Molar mass BaO2

    Moles BaO2 = 92.5 / 169.33 g/mol

    Moles BaO2 = 0.546 moles

    Step 4: Calculate moles of H2SO4

    For 1 mole BaO2 we need 1 mole of H2SO4

    For 0.546 moles of BaO2 we need 0.546 moles of H2SO4

    Step 5: Calculate the volume of H2SO4 needed

    Volume = moles / molarity

    Volume = 0.546 mol / 3.00 mol/L

    Volume = 0.182 L = 182 mL

    There is 182 mL of 3.00 M H2SO4 needed
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