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2 February, 10:36

In the following reaction, 451.4 g of lead reacts with excess oxygen forming 365.0 g of lead (II) oxide. Calculate the percent yield of the reaction.

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  1. 2 February, 10:50
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    Let MM (x) be the molar mass of x.

    MM (Pb) : MM (PbO)

    =207.21 : 223.20 = 451.4 g : x g

    cross multiply and solve for x

    x=223.2/207.21*451.4

    = 486.23 g

    Percentage yield = 365.0/486.23 = 0.75067 = 75.07% (rounded to 4 sign. fig.)
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