Balance the following equations;

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

Answer

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)

or

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))

or

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.