The rock in a lead ore deposit contains 84% PbS by mass. How many kilograms of the rock must be processed to obtain 1.1 kg of Pb?
i need to know what the conversion factors look like and how u did it
To solve this problem, we need to set up a conversion using the information provided and the concept of percent composition.
Step 1: Understand the information given
- The rock in the lead ore deposit contains 84% PbS (lead sulfide) by mass.
- We want to know how many kilograms of the rock must be processed to obtain 1.1 kg of pure Pb (lead).
- The conversion factor we need is the mass ratio of Pb to PbS in the rock.
Step 2: Set up the conversion factor
- Since the rock contains 84% PbS by mass, we know that the remaining 16% must be other constituents.
- Since we are interested in Pb, we want to find the mass of Pb in 100% of the rock.
- The mass of Pb in the rock can be calculated as 16% of the total rock mass: 1.1 kg * (16/100) = 0.176 kg.
Step 3: Convert from PbS to Pb
- To convert from PbS to Pb, we need to determine the ratio of their molar masses. The molar mass of PbS is 207.2 g/mol for lead (Pb) and 32.1 g/mol for sulfur (S).
- The molar mass ratio is calculated as 207.2 g/mol / 239.3 g/mol (Pb / PbS). This gives us a conversion factor of 0.8659.
Step 4: Calculate the mass of rock to be processed
- Now, we can use the conversion factor to determine how much rock must be processed to obtain 0.176 kg of Pb.
- We divide the mass of Pb by the conversion factor: 0.176 kg / 0.8659 = 0.203 kg.
Therefore, to obtain 1.1 kg of Pb, approximately 0.203 kg of the rock must be processed.
The key conversion factors used in this calculation were the percent composition of the rock (84% PbS) and the molar mass ratio of Pb to PbS (0.8659).