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26 December, 14:33

C. If 62.9 g of lead (II) chloride is produced, how many grams of lead (II) nitrate were

reacted?

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Answers (1)
  1. 26 December, 14:51
    0
    Mass = 76.176 g

    Explanation:

    Given dа ta:

    Mass of lead (II) chloride produced = 62.9 g

    Mass of lead (II) nitrate used = ?

    Solution:

    Chemical equation:

    Pb (NO₃) ₂ + 2HCl → PbCl₂ + 2HNO₃

    Number of moles of lead (II) chloride:

    Number of moles = mass/molar mass

    Number of moles = 62.9 g / 278.1 g/mol

    Number of moles = 0.23 mol

    Now we will compare the moles of lead (II) chloride with Pb (NO₃) ₂ from balance chemical equation:

    PbCl₂ : Pb (NO₃) ₂

    1 : 1

    0.23 : 0.23

    Mass of Pb (NO₃) ₂:

    Mass = number of moles * molar mass

    Mass = 0.23 mol * 331.2 g/mol

    Mass = 76.176 g
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