Question

The rock in a lead ore deposit contains 89 % PbS by mass. How many kilograms of the rock must be processed to obtain 1.5 kg of Pb?

Answer 1

Answer:

Approximately 1.9 kilograms of this rock.

Explanation:

Relative atomic mass data from a modern periodic table:

• Pb: 207.2;
• S: 32.06.

To answer this question, start by finding the mass of Pb in each kilogram of this rock.

89% of the rock is $\rm PbS$. There will be 890 grams of $\rm PbS$ in one kilogram of this rock.

Formula mass of $\rm PbS$:

$M(\mathrm{PbS}) = 207.2 + 32.06 = 239.26\; g\cdot mol^{-1}$.

How many moles of $\rm PbS$ formula units in that 890 grams of $\rm PbS$?

$\displaystyle n = \frac{m}{M} = \rm \frac{890}{239.26} = 3.71980\; mol$.

There's one mole of $\rm Pb$ in each mole of $\rm PbS$. There are thus $\rm 3.71980\; mol$ of $\rm Pb$ in one kilogram of this rock.

What will be the mass of that $\rm 3.71980\; mol$ of $\rm Pb$?

$m(\mathrm{Pb}) = n(\mathrm{Pb}) \cdot M(\mathrm{Pb}) = \rm 3.71980 \times 207.2 = 770.743\; g = 0.770743\; kg$.

In other words, the $\rm PbS$ in 1 kilogram of this rock contains $\rm 0.770743\; kg$ of lead $\rm Pb$.

How many kilograms of the rock will contain enough $\rm PbS$ to provide 1.5 kilogram of $\rm Pb$?

$\displaystyle \frac{1.5}{0.770743} \approx \rm 1.9\; kg$.