The rock in a lead ore deposit contains 84% PbS by mass. How many kilograms of the rock must be processed to obtain 1.0 kg of Pb?

1 kg Pb = ? mols Pb. mols = grams/molar mass

1 mol PbS = 1 mol Pb
kg PbS required = mols PbS x molar mass PbS.

mass PbS/0.84 = mass lead ore deposit that must be processed.
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To solve this problem, we need to determine the mass of the lead (Pb) in the rock and then find out how much rock is needed to obtain 1.0 kg of Pb.

First, let's assume we have x kilograms of the rock.

We are given that the rock contains 84% PbS by mass. This means that 84% of the rock is PbS and the rest is other components.

To find the mass of Pb in the rock, we need to calculate 84% of x:

Mass of Pb in the rock = 0.84x kilograms

Now, we want to find out how much rock is needed to obtain 1.0 kg of Pb. Since the ratio of Pb in the rock is 0.84, we can set up the following equation:

Mass of Pb in the rock / Mass of rock = Mass of Pb / 1.0 kg

Using the values we have:

0.84x / x = 1.0 kg

Now we can solve for x:

0.84x = 1.0 kg

Dividing both sides of the equation by 0.84:

x = 1.0 kg / 0.84

x ≈ 1.19 kg

Therefore, approximately 1.19 kg of the rock must be processed to obtain 1.0 kg of Pb.