One half of a balanced chemical equation is shown. 3Mg(OH)2 + 2H3PO4 Which lists the numbers of each atom the other half of the
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<u>Answer:</u> The mass percent of lead in lead (IV) carbonate is 63.32 %
<u>Explanation:</u>
The given chemical formula of lead (IV) carbonate is
To calculate the mass percentage of lead in lead (IV) carbonate, we use the equation:
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:
Hence, the mass percent of lead in lead (IV) carbonate is 63.32 %