Question #afa55

According to the balanced chemical equation given to you

#"Pb"_ ((s)) + "PbO"_ (2(s)) + 2"H"_ 2"SO"_ (4(aq)) -> 2"PbSO"_ (4(s)) + 2"H"_ 2"O"_ ((l))#

every #1# mole of lead(IV) oxide, #"PbO"_2#, that takes part in the reaction produces #2# moles of lead(II) sulfate, #"PbSO"_4#.

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

#"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 #"606.52 g"# of lead(II) sulfate for every #"73.78 g"# of lead(IV) oxide that take part in the reaction.

This means that your sample will produce

#23.6 color(red)(cancel(color(black)("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.