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26 January, 05:33

Manganese (II) oxide, lead (IV) oxide, and nitric acid react to produce permanganic acid, lead (II) nitrate, and water according to the reaction above. How many moles of each product could be produced by the reaction of 8.90 moles of nitric acid with excess manganese (II) oxide and excess lead (IV) oxide

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  1. 26 January, 06:02
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    There will be produced:

    2.97 moles HMnO4

    4.45 moles Pb (NO3) 2

    2.97 moles H2O

    Explanation:

    Step 1: Data given

    Manganese (II) oxide = MnO2

    lead (IV) oxide = PbO2

    nitric acid = HNO3

    Moles of HNO3 = 8.90 moles

    Step 2: The balanced equation

    2MnO2 + 3PbO2 + 6HNO3 → 2HMnO4 + 3Pb (NO3) 2 + 2H2O

    Step 3: Calculate moles of reactants and products

    For 2 moles MnO2 we need 3 moles PbO2 and 6 moles HNO3 to produce 2 moles HMnO4, 3 moles Pb (NO3) 2 and 2 moles of water

    For 8.90 moles of HNO3, there will react:

    8.90 / 3 = 2.97 moles MnO2

    8.90 / 2 = 4.45 moles PbO2

    There will be produced:

    8.90/3 = 2.97 moles HMnO4

    8.90/2 = 4.45 moles Pb (NO3) 2

    8.90 / 3 = 2.97 moles H2O
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