# Question #afa55

Mar 11, 2017

${\text{194 g PbSO}}_{4}$

#### Explanation:

According to the balanced chemical equation given to you

${\text{Pb"_ ((s)) + "PbO"_ (2(s)) + 2"H"_ 2"SO"_ (4(aq)) -> 2"PbSO"_ (4(s)) + 2"H"_ 2"O}}_{\left(l\right)}$

every $1$ mole of lead(IV) oxide, ${\text{PbO}}_{2}$, that takes part in the reaction produces $2$ moles of lead(II) sulfate, ${\text{PbSO}}_{4}$.

You can actually convert this mole ratio to a gram ratio by using the molar masses of two compounds

${\text{1 mole PbO"_2/"2 moles PbSO"_4 = (1 color(red)(cancel(color(black)("mole PbO"_2))) * "73.78 g"/(1color(red)(cancel(color(black)("mole PbO"_2)))))/(2color(red)(cancel(color(black)("moles PbSO"_4))) * "303.26 g"/(1color(red)(cancel(color(black)("mole PbSO"_4))))) = "73.78 g PbO"_2/"606.52 g PbSO}}_{4}$

You now know that the reaction produces $\text{606.52 g}$ of lead(II) sulfate for every $\text{73.78 g}$ of lead(IV) oxide that take part in the reaction.

This means that your sample will produce

$23.6 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g PbO"_2))) * "606.52 g PbSO"_4/(73.78 color(red)(cancel(color(black)("g PbO"_2)))) = color(darkgreen)(ul(color(black)("194 g PbSO}}_{4}}}}$

The answer is rounded to three sig figs.