Question

1.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
Balance the equation and describe in words the electron transfer(s) that takes place.

Answer 1

Do you know how to balance it?

Answer 2

no

Answer 3

Ok 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 PbO
PbS + O2 --> 2PbO + SO2
Now Pb is unbalanced as well to balance I'll put a 2 in front of the PbS
2PbS + O2 --> 2PbO + SO2
Pb is balanced but S isn't anymore so we will put a 2 in front of the SO2
2PbS + O2 --> 2PbO + 2SO2
Now Pb is balanced and S is balanced only O is unbalanced
We have 2 on the left and 6 on the right
To balance we need 6 oxygen on the left
6/2=3
2PbS + 3O2 --> 2PbO + 2SO2
That is the balanced equation

Answer 4

okay