At temperatures above 930 ∘C, PbO reduces PbS to form metallic lead (Pb). The byproduct of the reaction is SO2. Calculate the amount of lead produced (in kg) when a reactor is charged with 887.0 kg of PbS and 1330 kg of PbO.

To calculate the amount of lead (Pb) produced, we need to determine the limiting reactant in the reaction between PbS and PbO. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.

Let's start by writing the balanced chemical equation for the reaction:

PbS + 2PbO → 3Pb + SO2

From the balanced equation, we can see that the stoichiometric ratio between PbS and Pb is 1:3. This means that for every 1 mole of PbS, we need 3 moles of PbO to completely react and produce 3 moles of Pb.

To calculate the moles of PbS and PbO, we can use their molar masses.

The molar mass of PbS is:
207.2 g/mol (atomic mass of Pb) + 32.1 g/mol (atomic mass of S) = 239.3 g/mol

The moles of PbS can be calculated by dividing the mass of PbS by its molar mass:
Moles of PbS = Mass of PbS / Molar mass of PbS

Given that 1 kg = 1000 g, the mass of PbS in kg is 887.0 kg, which is equal to 887,000 g.

Moles of PbS = 887,000 g / 239.3 g/mol

Next, we can calculate the moles of PbO in a similar manner.

The molar mass of PbO is:
207.2 g/mol (atomic mass of Pb) + 16.0 g/mol (atomic mass of O) = 223.2 g/mol

The mass of PbO in kg is 1330 kg, which is equal to 1,330,000 g.

Moles of PbO = 1,330,000 g / 223.2 g/mol

Now we need to compare the moles of PbS and PbO to determine the limiting reactant. The limiting reactant is the one with the smaller number of moles because it would be consumed first during the reaction.

Let's calculate the moles of Pb that can be produced from each reactant:

Moles of Pb from PbS = Moles of PbS × (3 moles of Pb / 1 mole of PbS)
Moles of Pb from PbO = Moles of PbO × (3 moles of Pb / 2 moles of PbO)

Finally, we can calculate the mass of Pb produced using the moles and the molar mass:

Mass of Pb produced = Moles of Pb × Molar mass of Pb

I will perform these calculations and provide you with the final answer.