Balance the following equations;

c) C4H10(g) + O2(g) ----> CO2(g) + H2O(l)


C4H10(g) + O2(g) ----> CO2(g) + H2O(l)

Since oxygen is an element we will balance it last. Another useful rule is to balance non-oxygen and non-hydrogen atoms first. In this case that means carbon. In the unbalanced equation there are 4 carbon atoms in the reactants and 1 in the products.

C4H10(g) + O2(g) ----> 4CO2(g) + H2O(l)

Now we can balance hydrogen. There are 10 hydrogen in the reactants and 2 in the products. So a coefficient of 5 is needed for water.

C4H10(g) + O2(g) ----> 4CO2(g) + 5H2O(l)

So only oxygen remains. There are 13 oxygen atom in the products and two in the reactants. We need 13 oxygen atom in the reactants, but how do we get 13 oxygen atoms on the reactants side? Thirteen oxygen atoms is the same as 6 and 1/2 O2 molecules.

C4H10(g) + 6 1/2O2(g) ----> 4CO2(g) + 5H2O(l)


C4H10(g) + 13/2O2(g) ----> 4CO2(g) + 5H2O(l)

We prefer to have whole number coefficients when balancing chemical equations. To adjust to whole numbers we will multiply all the coefficients by 2.

2 x (C4H10(g) + 6 1/2O2(g) ----> 4CO2(g) + 5H2O(l))


2C4H10(g) + 13O2(g) ----> 8CO2(g) + 10H2O(l)

Notice all the atoms are now balanced. There are 8 C's, 20 H's and 26 O's on both sides of the equation.