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.