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?

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massPb=Percent ore * percentleadinPbS *massore

so what is the percent of lead in the PbS
= 207/(207+32)= you do it.

mass ore needed=2.3kg/(.84*percentleadinPbS)

To solve this problem, we need to determine the amount of lead (Pb) in the rock and then calculate how many kilograms of the rock need to be processed to obtain 2.3 kg of Pb.

Step 1: Calculate the mass of lead in the rock.
Since the rock contains 84% PbS by mass, we can assume that 84% of the rock's mass is made up of lead.
Let's assume the mass of the rock is "m" kg.
So, the mass of lead in the rock is 84% of "m" kg, which is (84/100) * m = 0.84m kg.

Step 2: Set up a ratio to determine the mass of lead required to obtain 2.3 kg of Pb.
From Step 1, we know that the mass of lead in the rock is 0.84m kg.
We are trying to obtain 2.3 kg of Pb.
So, we can set up the following ratio:
0.84m kg / 2.3 kg = m kg / x, where "x" is the mass of the rock that needs to be processed.

Step 3: Solve for "x".
To isolate "x", we can cross-multiply and then divide both sides of the equation:
(0.84m kg) * x = (2.3 kg) * m kg,
0.84x = 2.3m,
x = 2.3m / 0.84.

Step 4: Calculate the value of "x".
Using the given values, we plug them into the equation:
x = 2.3m / 0.84 = 2.738m kg.

Therefore, to obtain 2.3 kg of Pb, approximately 2.738 times the mass of the rock needs to be processed.