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dimulka [17.4K]
3 years ago
8

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

eld of the reaction.
Chemistry
1 answer:
user100 [1]3 years ago
4 0
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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Boyle's law is expressed as;

P₁V₁ = P₂V₂

Where P₁ is Initial Pressure, V₁ is Initial volume, P₂ is Final Pressure and V₂ is Final volume.

Given that;

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Learn more about Boyle's law here: brainly.com/question/1437490

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