Balance the following equations;

b) C4H9SH(g) + O2(g) --heat--> CO2(g) + H2O(g) + SO2(g)

Answer

This equation appears to be a little more complicated that the formation equations we were discussing. But if maintain control we'll see these are not that difficult. It is all in the approach.

To balance combustion reactions always balance the element oxygen, O2, last. If we do that we need to balance carbon, hydrogen and sulfur first, we'll do that beginning with carbon.

C4H9SH(g) + O2(g) --heat--> CO2(g) + H2O(g) + SO2(g)

There are four carbon atoms (C4H9SH) on the left side and one carbon atom (CO2) on the right side. Changing the coefficient before CO2 will balance the carbon atoms. The new coefficent is '4'.

C4H9SH(g) + O2(g) --heat--> 4CO2(g) + H2O(g) + SO2(g)

Now hydrogen...let's see ten hydrogen atoms on the left (C4H9SH) and only two on the right (H2O). So I'll need to change the coefficient before H2O from '1' to '5' and that will balance the hydrogen atoms.

C4H9SH(g) + O2(g) --heat--> 4CO2(g) + 5H2O(g) + SO2(g)

To balance sulfur, there is one sulfur atom on the left and , hey there is also one sulfur atom on the right. Alright! the sulfur is balanced!

Now we'll balance the oxygen atoms. There are two oxygen atoms (O2) on the left and fifteen oxygen atoms on the right (4CO2 + 5H2O + SO2). So we need fifteen oxygen atoms on the left. But on the left oxygen only comes in pairs. How do we get fifteen oxygen atoms, we can get fourteen by placing a '7' in front of O2, or we can get sixteen by placing a coefficent of '8' in front of O2. How do we get fifteen? We can use a coefficient of 7.5, or 15/2 (fifteen halves).

C4H9SH(g) + 15/2O2(g) --heat--> 4CO2(g) + 5H2O(g) + SO2(g)

 

But we do not want fractional coefficients, so we'll multiply the complete equation by 2!

2 (C4H9SH(g) + 15/2O2(g) --heat--> 4CO2(g) + 5H2O(g) + SO2(g))

and we get,

2C4H9SH(g) + 15O2(g) --heat--> 8CO2(g) + 10H2O(g) + 2SO2(g))

Now the equation is balanced.