Question

In the following reaction, 451.4 g of lead reacts with excess oxygen forming 321.3 g of lead(II) oxide. Calculate the percent yield of the reaction.

Answer 1

yuck percent yield.
first things first write out a balanced equation.
then find the theoretical yield:
\[\frac{451.4g lead}{1}*\frac{1mole-lead}{grams-lead}*\frac{moles-lead(II) oxide}{moles-lead}(*ratio*)\]
\[*\frac{grams-lead(II)oxide}{1mole-lead(II)oxiide}=theoretical yield\]
then:
\[percent.yield=\frac{321.3 g}{theoretical yield}*100\]