Question 3.EP.1: Propane, C3H8, is used as a fuel for gas barbecue grills, wh...
Propane, C_{3}H_{8}, is used as a fuel for gas barbecue grills, where it burns in a controlled fashion. But if a mixture of propane and air is ignited in a closed space, like a gas pipeline, an explosion can easily result. In either of these cases, the propane combines with oxygen, O_{2}, to form carbon dioxide and water. Write a balanced chemical equation describing this reaction.
Strategy
This problem requires two steps. First we must read the problem and determine which substances are reactants and which are products. This will allow us to write an unbalanced “skeleton” equation. Then we must proceed to balance the equation, making sure that the same number of atoms of each element appears on both sides.
Step 1: The problem notes that propane burns or explodes by combining with oxygen. The reactants are therefore C_{3}H_{8} and O_{2}. We are also told that the products are CO_{2} and H_{2}O. So we have enough information to write the unbalanced equation. (Here we will write “blanks” in front of each formula to emphasize that we still need to determine the coefficients.)
—C_{3}H_{8} + —O_{2} \longrightarrow —CO_{2}+ —H_{2}O
Step 2: Balancing an equation like this is aided by making some observations. In this case, both carbon and hydrogen appear in only one place on each side of the equation. Our first steps will be to balance these two elements, because there will be no other way to adjust them. Let’s begin with carbon. (This choice is arbitrary.) To achieve the required balance, we can use stoichiometric coefficients of one for C_{3}H_{8} and three for CO_{2}, giving three carbon atoms on each side of the equation. At this point, we have
1 \ C_{3}H_{8} + —O_{2} \longrightarrow 3 \ CO_{2}+ —H_{2}O
(The coefficient of one in front of the propane would normally be omitted, but is explicitly written here for emphasis.) Next we will balance hydrogen. The propane already has its coefficient set from our work on carbon, so it dictates that there are eight H atoms to be accounted for by water on the product side of the equation. We need to insert a coefficient of four to achieve this balance.
1 \ C_{3}H_{8} + —O_{2} \longrightarrow 3 \ CO_{2}+ 4 \ H_{2}O
This leaves us with only oxygen to balance. We have saved this for last because it appears in three compounds rather than two and also because it appears as an element rather than a compound on the left side of the equation, which makes it easy to adjust the number of oxygen atoms without upsetting the balance of the other elements. Because there are ten O atoms on the product side of the equation, we need ten on the reactant side. We can easily achieve this by using 10/2 = 5 for the stoichiometric coefficient of the O_{2} reactant. This gives us the balanced chemical equation,Molecules, Moles, and Chemical Equations
1 \ C_{3}H_{8} + 5 \ O_{2} \longrightarrow 3 \ CO_{2}+ 4 \ H_{2}O
Normally, this would be written without showing the coefficient of one in front of the propane:
C_{3}H_{8} + 5 \ O_{2} \longrightarrow 3 \ CO_{2}+ 4 \ H_{2}O
Analyze Your Answer
Our final equation has three carbon atoms, eight hydrogen atoms, and ten oxygen atoms on each side, so it is balanced.