How many grams of lead (ll) sulfate would be produced from the complete reaction of 23.6 g lead (iV) oxide ?

29.92grams of PbSO4

Explanation:

lead (iV) oxide = PbO2 = Molar mass: 239.2 g/mol

lead (ll) sulfate = PbSO4 = Molar mass: 303.26 g/mol

PbO2 = PbSO4

1:1 ratio

Pb = Lead

Lead has an oxidation number of 4+

O = Oxygen

Oxygen has an oxidation number of 2-

PbO2 + 4H+ + SO4 2- + 2e- = PbSO4(s) + 2H2O

Ok so the above would be the likely complete reaction, though we don't really need this as we already know the ratio is 1:1.

23.6g of PbO2

23.6/239.2 = 0.09866 Moles of PbO2

Since we have a 1:1 ratio we know that the same number of moles of PbSO4 are produced and since we know the molar mass it's simply molar mass multiplied by number of moles.

303.26 x 0.09866 = 29.92grams of PbSO4

output?

explanation: