Question

How many atoms of Lead are in a 62.7 g sample of Lead(II) permanganate?

Answer 1

Answer:

$\large \boxed {8.48 \times 10^{22}\text{ atoms}}$

Explanation:

1 mol of Pb(MnO₄)₂contains 1 mol of Pb atoms.

1. Moles of Pb

$\text{Moles of Pb} = \text{62.7 g Pb(MnO _{4} )}_{2} \times \dfrac{\text{1 mol Pb(MnO _{4} )}_{2}}{\text{445.07 g Pb(MnO _{4} )}_{2}}\\\\\times \dfrac{\text{1 mol Pb}}{\text{1 mol Pb(MnO _{4} )}_{2}} = \text{0.1409 mol Pb}$

2. Atoms of Pb

$\text{Atoms of Pb} = \text{0.1409 mol Pb } \times \dfrac{ 6.022 \times 10^{23}\text{ atoms Pb }}{\text{1 mol Pb }}\\\\= \large \boxed {\mathbf{8.48 \times 10^{22}}\textbf{ atoms Pb }}$