Answer:
The correct answer is d.
Explanation:
Step 1: First we must obtain the molecular weight of [tex]PbCO_{3}[/tex] and Pb.
[tex]PM_{PbCO_{3} }[/tex] = 267,21 [tex]\frac{g}{mol}[/tex]
[tex]PM_{Pb}[/tex] = 207,2 [tex]\frac{g}{mol}[/tex]
Step 2: We can now calculate the percentage of lead in a [tex]PbCO_{3}[/tex] molecule by stating a simple rule of three:
[tex]PM_{PbCO_{3} }[/tex] --> 100%.
[tex]PM_{Pb}[/tex] --> X
%Pb = [tex]\frac{PM_{Pb}}{ PM_{PbCO_{3} }} * 100%[/tex]
%Pb = [tex]\frac{207.2 \frac{g}{mol}}{267.21 \frac{g}{mol}} * 100%[/tex]
%Pb = 77,54 %
Have a nice day!
