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The following unbalanced equation describes the reaction that can occur when lead (II) sulfide reacts with oxygen gas to produce lead (II) oxide and sulfur dioxide gas: PbS + O2  PbO + SO2.

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 so the only one that is currently unbalanced is the oxygen there is two on the left and three on the right side If we add a two in front of the PbOPbS + O2 --> 2PbO + SO2Now Pb is unbalanced as well to balance I'll put a 2 in front of the PbS2PbS + O2 --> 2PbO + SO2Pb is balanced but S isn't anymore so we will put a 2 in front of the SO22PbS + O2 --> 2PbO + 2SO2Now Pb is balanced and S is balanced only O is unbalanced We have 2 on the left and 6 on the rightTo balance we need 6 oxygen on the left6/2=32PbS + 3O2 --> 2PbO + 2SO2 That is the balanced equation

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