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25 July, 21:17

How many grams of lead (II) chloride would be produced by reacting 50.0g of sodium chloride

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  1. 25 July, 21:46
    0
    =302.28 grams

    Explanation:

    Lead (II) Chloride is produced by the reaction between the reaction between Pb (NO₃) ₂ and NaCl according to the equation below.

    Pb (NO₃) ₂ + 2NaCl → PbCl₂ + 2 NaNO₃

    From the equation 2 moles of NaCl produce 1 mole of PbCl₂

    Therefore the reaction ratio is 2:1

    In 50.0 g of sodium, there are 50/23 moles

    =2.174 moles

    Since the reaction to product ratio is 2:1, then the number of moles of PbCl₂ produced is. 2.174 moles*1/2=1.087 moles.

    Mass = number of moles * RMM

    =1.087 moles*278.1grams/mol

    =302.28 grams
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