Butane burns completely in oxygen according to the equation 2C4H10 plus 13O2 reacting to give 8CO2 plus 10H2O. If 44 grams of carbon dioxide is produced in this reaction, how much oxygen is consumed?
In this question, we have butane, C4H10, which is a gas under standard conditions, reacting with gaseous oxygen. Burning, properly known as combustion, is a reaction with oxygen which produces much energy in the form of heat or light. So we could write plus energy at the end of the equation. Because energy is released in the reaction, we know this is an exothermic reaction. Both the products, carbon dioxide and water, are given off in the gaseous form.
We are told that butane burns completely, which means that all of the butane is used up in the reaction. Let’s now look at the data that we are given. We are told that 44 grams of carbon dioxide is produced. So here I have written the mass of carbon dioxide of 44 grams underneath CO2. And we are asked to find how much oxygen is consumed.
Usually, the term “how much” refers to mass. So under O2, I’ve written 𝑚 equals question mark. This is all the data that we have been given. We will need to use the data given for CO2 to determine the data needed for O2. However, we cannot directly relate mass values. It is the stoichiometric coefficients which we can relate which allow us to relate the number of moles.
So let’s begin by working out the number of moles of CO2 that is equivalent to 44 grams. One of the ways to calculate this is to use the equation number of moles is equal to mass over molar mass. We know the mass of CO2 produced, 44 grams. And we need to calculate the molar mass of CO2. The molar mass for carbon is 12.011 grams and for oxygen 15.999 grams, which we will need to multiply by two because there are two oxygen atoms present in CO2. And we get an answer of 44.009 grams per mole. We can put this value into our equation for the molar mass of carbon dioxide, which gives us 0.999795 moles. Let’s keep the decimal places on our calculator for now and round off at the very end.
Now, we can relate the number of moles of CO2 that we just calculated with the number of moles of O2 consumed. According to the balanced equation, 13 moles of O2 produce eight moles of CO2. So 𝑥 moles of O2 will produce 0.999795 moles of CO2. If we solve for 𝑥, we will take 13 multiplied by the number of moles of CO2 on our calculator divided by eight. And we get a value for the number of moles of oxygen consumed of 1.624666 moles.
The last step in this calculation is to convert the number of moles of O2 into a mass value. Let’s clear some space to do this. The mass of O2 consumed is equal to the number of moles times by the molar mass. We can put in the number of moles we have just calculated. And we will quickly have to first work out the molar mass of O2.
There are two oxygen atoms, each with a molar mass of 15.999 grams per mole, which gives a molar mass for O2 of 31.998 grams per mole. We can put this value into our equation. And we get 51.986062 grams of O2, which we can write, to three decimal places, 51.986 grams.
The question did not specify how many decimal places or significant figures we should use. So I have chosen three decimal places. If we chose to follow the correct number of significant figures, we would have to use two significant figures because the only piece of data we are given, the 44 grams of carbon dioxide, is expressed to two significant figures. And if you did this, you’d get 52 grams as the mass of oxygen consumed.
Finally, if 44 grams of carbon dioxide is produced, how much oxygen is consumed? And the answer is 52 grams.