Answer:
Approximately 1.9 kilograms of this rock.
Explanation:
Relative atomic mass data from a modern periodic table:
To answer this question, start by finding the mass of Pb in each kilogram of this rock.
89% of the rock is [tex]\rm PbS[/tex]. There will be 890 grams of [tex]\rm PbS[/tex] in one kilogram of this rock.
Formula mass of [tex]\rm PbS[/tex]:
[tex]M(\mathrm{PbS}) = 207.2 + 32.06 = 239.26\; g\cdot mol^{-1}[/tex].
How many moles of [tex]\rm PbS[/tex] formula units in that 890 grams of [tex]\rm PbS[/tex]?
[tex]\displaystyle n = \frac{m}{M} = \rm \frac{890}{239.26} = 3.71980\; mol[/tex].
There's one mole of [tex]\rm Pb[/tex] in each mole of [tex]\rm PbS[/tex]. There are thus [tex]\rm 3.71980\; mol[/tex] of [tex]\rm Pb[/tex] in one kilogram of this rock.
What will be the mass of that [tex]\rm 3.71980\; mol[/tex] of [tex]\rm Pb[/tex]?
[tex]m(\mathrm{Pb}) = n(\mathrm{Pb}) \cdot M(\mathrm{Pb}) = \rm 3.71980 \times 207.2 = 770.743\; g = 0.770743\; kg[/tex].
In other words, the [tex]\rm PbS[/tex] in 1 kilogram of this rock contains [tex]\rm 0.770743\; kg[/tex] of lead [tex]\rm Pb[/tex].
How many kilograms of the rock will contain enough [tex]\rm PbS[/tex] to provide 1.5 kilogram of [tex]\rm Pb[/tex]?
[tex]\displaystyle \frac{1.5}{0.770743} \approx \rm 1.9\; kg[/tex].
