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11 January, 02:57

An industrial vat contains 650 grams of solid lead (II) chloride formed from a reaction of 870 grams of lead (II) nitrate with excess hydrochloric acid. This is the equation of the reaction: 2HCl + Pb (NO3) 2 → 2HNO3 + PbCl2. What is the percent yield of lead (II) chloride? The percent yield of lead chloride is %

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Answers (2)
  1. 11 January, 03:17
    0
    89%

    Explanation:

    Round it up
  2. 11 January, 03:25
    0
    88.98 %

    Solution:

    The Balance Chemical Equation is as follow,

    2 HCl + Pb (NO₃) ₂ → 2 HNO₃ + PbCl₂

    According to equation,

    331.2 g (1 mole) Pb (NO₃) ₂ produces = 278.1 g (1 mole) PbCl₂

    So,

    870 g of Pb (NO₃) ₂ will produce = X g of PbCl₂

    Solving for X,

    X = (870 g * 278.1 g) : 331.2 g

    X = 730.5 g of PbCl₂

    Therefore,

    Theoretical Yield = 730.5 g

    Also as given,

    Actual Yield = 650 g

    So using following formula for percentage yield,

    %age Yield = (Actual Yield / Theoretical Yield) * 100

    Putting values,

    %age Yield = (650 g / 730.5 g) * 100

    %age Yield = 88.98 %
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