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 807.0 kg of PbS and 1210 kg of PbO
Molar mass of PbO: 446.4 g/mol
Molar mass of Pb: 621.6 g/mol
(1210000g/446.4)*621.6 = 1684892g = 1684 kg
Jasmin, Thank you for the explanation.
To calculate the amount of lead produced in this reaction, we first need to determine the limiting reagent. The limiting reagent is the reactant that will be completely consumed first, thus limiting the amount of product that can be formed.
Let's start by writing the balanced chemical equation for the reaction:
PbO + PbS → 2Pb + SO2
From the balanced equation, we can see that the stoichiometric ratio between PbO and PbS is 1:1, meaning one mole of PbO reacts with one mole of PbS.
Next, we need to convert the given masses of PbS and PbO to moles. To do this, we will use the molar masses of the compounds:
Molar mass of PbS = 207.2 g/mol
Molar mass of PbO = 223.2 g/mol
Converting the masses to moles:
Moles of PbS = 807.0 kg * (1000 g/kg) / (207.2 g/mol)
Moles of PbO = 1210 kg * (1000 g/kg) / (223.2 g/mol)
Now, we compare the number of moles of each reactant to determine the limiting reagent.
Moles of PbS / Moles of PbO = (807.0 kg * (1000 g/kg) / (207.2 g/mol)) / (1210 kg * (1000 g/kg) / (223.2 g/mol))
Simplifying the expression:
Moles of PbS / Moles of PbO = (807.0 g * 223.2 g/mol) / (1210 g * 207.2 g/mol)
Moles of PbS / Moles of PbO ≈ 3.195
Since the ratio is approximately 3.195, we can conclude that PbS is the limiting reagent. This means that all the PbS will react, and PbO will be in excess.
Now, we need to calculate the moles of Pb produced. From the balanced equation, we can see that for every 1 mole of PbS, 2 moles of Pb are produced.
Moles of Pb = 2 * Moles of PbS
Finally, we convert the moles of Pb to kilograms:
Mass of Pb = Moles of Pb * Molar mass of Pb
Mass of Pb = (2 * Moles of PbS) * Molar mass of Pb
Mass of Pb = (2 * (807.0 kg * (1000 g/kg) / (207.2 g/mol))) * 207.2 g/mol
Mass of Pb ≈ 7849.8 kg
Therefore, approximately 7849.8 kg of lead will be produced in the reaction.