Question

1. Using the acidic reaction below:

Pb_(s) + PbO_2(s) + SO₄^2-_(aq) -> PbSO_4(s)

Calculate the number of moles of lead required to produce 974.39 grams of PbSO_4(s) .

2. Pb_(s) + PbO_2(s) + SO₄^2-_(aq) -> PbSO_4(s)

Calculate the number of moles of PbSO_4(s) produced if 254.15 grams of Pb and 182.88 grams of PbO_2(s) were reacted.

Answer 1

1)
number of mole of PbSO4 produce = (given mass)/(molar mass of PbSO4)
molar mass of PbSO4 = 303.3 g/mol
number of mole of PbSO4 produce = 974.39/303.3
= 3.2 mole
reaction taking pkace is
Pb(s) + PbO2 (s) + SO42-(aq) -> PbSO4(s)
according to reaction
no of mole PbSO4 produce = number of mole of Pb used
so, number of mole of Pb required = 3.2 mole

Answer : 3.2 mole

2)
molar mass of Pb = 207.2 g/mol
molar mass of PbO2 = 239.2 g/mol
number of mole = (given mass)/(molar mass)
number of mole Pb = 182.88/207.2
= 0.88 mole
number of mole of PbO2 = 182.88/239.2
= 0.76 mole
Pb(s) + PbO2 (s) + SO42-(aq) -> PbSO4(s)
according to reaction given
number of mole Pb used = number of mole of PbO2 used
so, PbO2 is limiting reagent
solve the problem by considering the amount of PbO2 used
and also,
number of mole of PbSO4 produce = number of mole of PbO2 used
number of mole of PbSO4 produce = 0.76 mole

Answer : 0.76 mole