Question

Please Help. I need to get this done.

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.

Answer 1

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