Determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen

Answer 1

Hello Kayleigh,

To answer this question, we need to think about setting up our balanced chemical equation so that we can easily convert between reactants and products using stoichiometry and dimensional analysis. In this case, the balanced chemical equation that we should have with the combustion of propane would be:

C3H8 (g) + 5 O2 (g) >>>> 3 CO2 (g) + 4 H2O (l)

From this balanced chemical equation, we see that there are 5 moles of oxygen reacting for every 1 mole of propane (C3H8), so we will use that in our balanced chemical equation. From here, we want to use dimensional analysis and stoichiometry to convert from units of each reactant to units of a product and see which is less. The reactant that produces the lesser amount of product is considered the limiting reactant.

We can convert to either of our products, H2O or CO2. As long as we are converting to the same product for both conversions from either reactant, we will find the right answer. For example, let's convert both of our amounts of reactants to amounts of CO2 and see which one results in less CO2. That reactant will be our limiting reactant.

Let's start with 30.0 g. We determine that the molar mass of propane is 44.1 grams, which we can use to convert grams of a substance to moles of that substance.

30.0 g C3H8 * 1 mole C3H8/44.1 grams C3H8 = 0.68 moles C3H8

Once we have determined the number of moles in 33.0 g C3H8, we use the balanced chemical equation and the mole ratio between C3H8 and CO2 from that equation to find the moles of CO2.

0.68 moles C3H8 * 3 moles CO2/1 mole C3H8 = 2.04 moles CO2

Now that we found moles of CO2 produced from 30.0 g C3H8, we want to use that same process using 75.0 g oxygen.

We start with 75.0 g oxygen and we can determine that the molar mass of oxygen is 32.0 grams, which again we can use to convert between grams and moles of a substance.

75.0 g O2 * 1 mole O2/32.0 g O2 = 2.34 moles O2.

From here, we use the balanced chemical equation and the mole ratio between O2 and CO2 from that equation to find the moles of CO2.

2.34 moles O2 * 3 moles CO2/5 moles O2 = 1.40 moles CO2.

Since 1.40 moles of CO2 produced from 75.0 g oxygen is less than the 2.04 moles of CO2 produced from 30.0 g propane, we can conclude that the oxygen is the limiting reactant here.