Answer: The mass percent of lead in lead (IV) carbonate is 63.32 %
Explanation:
The given chemical formula of lead (IV) carbonate is [tex]Pb(CO_3)_2[/tex]
To calculate the mass percentage of lead in lead (IV) carbonate, we use the equation:
[tex]\text{Mass percent of lead}=\frac{\text{Mass of lead}}{\text{Mass of lead (IV) carbonate}}\times 100[/tex]
Mass of lead = (1 × 207.2) = 207.2 g
Mass of lead (IV) carbonate = [(1 × 207.2) + (2 × 12) + (6 × 16)] = 327.2 g
Putting values in above equation, we get:
[tex]\text{Mass percent of lead}=\frac{207.2g}{327.2g}\times 0100=63.32\%[/tex]
Hence, the mass percent of lead in lead (IV) carbonate is 63.32 %