Balance the following equation showing complete combustion of a hydrocarbon: C3H8 + O2 --> CO2 +H2O

First, we can look at the number of hydrogens on either side. The equation shows 8 hydrogen atoms on the left hand side (in C3H8) whereas on the right hand side, we have 2 hydrogen atoms (in H2O). So we need more on the right hand side; we already have 2 hydrogen atoms, so to get to 8, we can put a 4 in front of the H2O as 4 x 2 =8. The equation is now C3H8 + O2 —> CO2 + 4H2O Now, we can look at the carbons; on the left hand side we have 3 carbons whereas on the right hand side one, so more carbon atoms are required on the right hand side. To get 3 carbon atoms on the right hand side, we can put a 3 in front of the CO2 as 3 x 1=3. The equation is now C3H8 + O2 —> 3CO2 + 4H2O. We’re now left with the oxygens. There is only 2 oxygen atoms on the left hand side. On the right hand side, we have 3CO2 molecules = 3 x 2 = 6 oxygen atoms and 4H2O molecules= 4 x 1 =4 oxygen atoms, giving us a total of 10 oxygen atoms. To get 10 oxygens on the left hand side, we multiply the O2 by 5 as 5 x 2=10. The final equations is C3H8 + 5O2 —> 3CO2 + 4H2O. Finally, we check that everything balances; 3 carbons on the left and right, 8 hydrogens on the left and right and 10 oxygens on the left. The equation is now balanced correctly.