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 553.0 kg of PbS and 830 kg of PbO.
Molar mass of PbO: 446.4 g/mol
Molar mass of Pb: 621.6 g/mol
(830000g/446.4)*621.6 = convert answer from g to kg
1211
1851
To calculate the amount of lead produced, we need to determine the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and determines the amount of product that can be formed.
First, we need to write and balance the chemical equation for the reaction:
PbO + PbS → Pb + SO2
The molar mass of PbS is 239.27 g/mol and the molar mass of PbO is 223.2 g/mol.
Next, let's calculate the number of moles of PbO and PbS:
Number of moles of PbO = mass of PbO / molar mass of PbO
= 830 kg / 223.2 g/mol
= 3.716 mol
Number of moles of PbS = mass of PbS / molar mass of PbS
= 553.0 kg / 239.27 g/mol
= 23.095 mol
According to the balanced equation, the stoichiometric ratio between PbO and PbS is 1:1. Therefore, one mole of PbS reacts with one mole of PbO to produce one mole of Pb.
Now, we need to determine the limiting reactant by comparing the number of moles of each reactant. The smaller number of moles corresponds to the limiting reactant.
In this case, PbO is the limiting reactant since it has fewer moles (3.716 mol) compared to PbS (23.095 mol).
Next, we need to calculate the number of moles of Pb produced, taking into account the limiting reactant:
Number of moles of Pb = 3.716 mol
Finally, we can calculate the mass of lead produced:
Mass of Pb = Number of moles of Pb × molar mass of Pb
= 3.716 mol × 223.2 g/mol
= 829.2512 g
Converting grams to kilograms:
Mass of Pb = 829.2512 g / 1000
= 0.8293 kg
Therefore, approximately 0.8293 kg of lead (Pb) will be produced when a reactor is charged with 553.0 kg of PbS and 830 kg of PbO.