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

Approximately 1.9 kilograms of this rock.

Step-by-step 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 = (m)/(M) = \rm (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 * 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 (1.5)/(0.770743) \approx \rm 1.9\; kg`.