Answer
$$10.8 \space kg$$
Work Step by Step
1. Calculate the total mass of the sphere: ** Lead has a density of $11.4 \space g/cm^3$. (See table 1.4, page 19)
Sphere volume = $\frac {4\pi r^3} 3$
$radius = 5.00 \space cm$
$V = \frac{4 \pi (5.00 \space cm)^3}{3} = 524 \space cm^3$
Mass = $ V \times \rho = 524 \space cm^3 \times \frac{11.4 \space g}{cm^3} = 5970 \space g$
2. Find the amount of that ore that is required to obtain this mass of lead.
$5970 \space g \space (Lead) \times \frac{100 \space \% (Galena)}{86.6 \space \% (Lead)} \times \frac{100 \space \% (Ore)}{68.5 \space \% (Galena)} = 1.0 \times 10^4 \space g (Ore)$
But, that calculation assumes 100% efficiency in the extraction, which is not true for this situation.
$1.0 \times 10^4 \space g (Ore) \times \frac{100 \space \% (Raw \space ore)}{92.5 \space \% (Ore)} = 1.08 \times 10^4 \space g (Raw \space ore) \times \frac{1 \space kg}{10^3 \space g} = 10.8 \space kg \space (Raw \space ore)$
