Question

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.

Answer 1

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.)