posted by T on .
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
pbo in gram/ 446.4 = ans *621.6 = final ans