HELP!!!
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 487.0 kg of PbS and 750 kg of PbO.
Looks like the reaction is
2PbO + PbS = 3Pb + SO2
487kg PbS = 487000/239.26 = 2035.44 moles
750kg PbO = 750000/223.20 = 3360.22 moles
Since each PbS needs two PbO, the PbO is the limiting reagent.
Each 2 moles of PbO produce 3 moles of Pb, so we end up with
3/2 * 3360.22mole * 207.2g/mol = 1044.36kg Pb
Its wrong
pbo in gram/ 446.4 = ans *621.6 = final ans
To calculate the amount of lead produced, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To find the limiting reactant, we can compare the moles of each reactant to the stoichiometric ratio of the balanced chemical equation.
The balanced chemical equation for the reaction between PbO and PbS is:
PbO + PbS → 2Pb + SO2
Let's calculate the moles of PbO and PbS:
Molar mass of PbO = 207.2 g/mol (given)
Molar mass of PbS = 239.3 g/mol (given)
Moles of PbO = mass of PbO / molar mass of PbO
= 750 kg * 1000 g/kg / 207.2 g/mol
Moles of PbS = mass of PbS / molar mass of PbS
= 487.0 kg * 1000 g/kg / 239.3 g/mol
Now we have the moles of PbO and PbS. By comparing the moles to the stoichiometric ratio, we can determine which reactant is limiting.
From the balanced equation, we can see that 1 mole of PbS reacts with 1 mole of PbO to form 2 moles of Pb.
Therefore, the stoichiometric ratio of PbS to PbO is 1:1.
Now, compare the moles of PbS to the moles of PbO. The reactant with fewer moles is the limiting reactant.
If the moles of PbS are greater than the moles of PbO, PbO is the limiting reactant. Otherwise, PbS is the limiting reactant.
Once we determine the limiting reactant, we can calculate the moles of Pb that will be formed. Since 1 mole of PbS produces 2 moles of Pb, the moles of Pb produced will be twice the moles of the limiting reactant.
Finally, to calculate the mass of Pb produced, we need to convert moles of Pb to mass using the molar mass of Pb.
Molar mass of Pb = 207.2 g/mol (given)
Mass of Pb = moles of Pb * molar mass of Pb
To convert the mass of Pb to kg, divide by 1000.
I hope this explanation helps you in calculating the amount of lead produced when the reactor is charged with PbS and PbO.