# How much lead can be extracted from 254.3 g of lead (IV) oxide?

Oct 29, 2016

Approx. $215 \cdot g$.
$\text{Moles of plumbic oxide}$ $=$ $\frac{254.3 \cdot g}{239.3 \cdot g \cdot m o {l}^{-} 1}$, which is slightly over a mole.
i.e.: $\frac{254.3 \cdot g}{239.3 \cdot g \cdot m o {l}^{-} 1} \times 207.2 \cdot g \cdot m o {l}^{-} 1$ $=$ ??g