The rock in a lead ore deposit contains 84% pbs by mass. how many kilograms of the rock must be processed to obtain 2.3 kg of pb?
kg

3.162 kg

Explanation:

1) Mass percent = (mass of a part / total mass) × 100

2) 84% PbS by mass.

Percent PbS = ( mass of pure PbS / mass of the rock) × 100 = 84%

⇒ mass of the rock = mass of pure PbS × 100 / 84 =

⇒ mass of the rock = mass of pure PbS / 0.84 ← this is fhe formula that you will uise at the end.

2) Determine the mass of pure PbS that contains 2.3 kg of Pb

i) Unit formula: PbS

ii) molar mass of PbS: 207.2 g/mol + 32.065 g/mol = 239.265 g/mol

iii) proportion:

207.2 g of Pb / 239.265 g of PbS = 2,300 g of Pb / x

⇒ x = 2,300 × 239.265 g of PbS / 207.2 = 2,655.93 g of PbS = 2.656 Kg of PbS

4) Use the formula to determine how many kilograms of the rock contain 2.565 kg of pure PbS

mass of the rock = mass of pure PbS / 0.84 = 2.565 / 0.84 = 3.162 kg

i believe the answer is true.