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 produced, we need to determine the limiting reactant in the reaction. The limiting reactant is the one that is completely consumed first and determines the amount of product formed.
Let's start by writing the balanced chemical equation for the reaction:
PbO + PbS → 2Pb + SO2
The molar mass of PbS is 239.3 g/mol, and the molar mass of PbO is 223.2 g/mol.
1. Calculate the number of moles of PbS and PbO:
Moles of PbS = mass of PbS / molar mass of PbS
= 887.0 kg / 239.3 g/mol
= 3705.47 mol
Moles of PbO = mass of PbO / molar mass of PbO
= 1330 kg / 223.2 g/mol
= 5954.80 mol
2. Determine the stoichiometric ratio between PbS and PbO using the balanced equation. From the equation, we can see that 1 mol of PbS reacts with 1 mol of PbO to produce 2 moles of Pb:
Ratio of moles of PbS to PbO = 1:1
3. Determine the limiting reactant. The reactant that has the smallest number of moles is the limiting reactant. In this case, both reactants have the same number of moles, so either one can be the limiting reactant.
4. Calculate the amount of lead produced using the limiting reactant. Since the stoichiometric ratio is 1 mol of PbS to 2 moles of Pb, we can calculate the number of moles of lead produced using the moles of the limiting reactant:
Moles of Pb produced = Moles of limiting reactant (PbS) = 3705.47 mol
5. Calculate the mass of lead produced:
Mass of Pb produced = Moles of Pb produced × Molar mass of Pb
= 3705.47 mol × 207.2 g/mol
= 767,042.58 g
= 767.04 kg
Therefore, when the reactor is charged with 887.0 kg of PbS and 1330 kg of PbO, the amount of lead produced is approximately 767.04 kg.