The number of moles and the mass of water formed by the combustion of 20.0 kg Octane C8H18, in an excess of oxygen.

Question

The number of moles and the mass of water formed by the combustion of 20.0 kg  

           Octane C₈H₁₈, in an excess of oxygen.  

That is the question, I am unsure of what I must do. A bit confused

Answer 1

Hello, Arianna,

First, we need a balanced equation. That is an explanation of its own, which I'd be glad to do, but there is limited space in the reply box, so I'll make it short. Please ask, if you want more details.

It is a little harder to balance than most combustion reactions because of the need for two C8H18 molecules to get everything balanced. I started with just one, but soon realized that the oxygen comes in pairs (O2) but is consumed as as a single (H2O). To make it work I would need 12 1/2 molecules of O2, which isn't permitted. I just multiplied everything by two to get the equation below. Long story short, I needed to increase the octane to 2 to deal with the even/odd oxygen count, and found that to work. Each side of the reaction now has the same number of H, C, and O

Next, let's calculate the number of moles of octane. We are given 20kg, which I'll change to 20,000 grams. The molar mass of octane is 114 g/mole (add the atomic masses of each element in the molecule: 8x12 for C and 1x18 for H, to yield 114). I have 20,000 g, so to calculate moles I divide 20,000g/(114g/mole). The grams cancel and moles moves to the top. We have 175 moles of octane (to 3 sig figs). Note how the units cancel out to give us the correct value.

We have an excess of oxygen, so the most we can get of the CO2 and H2O) is if all the octane combusts. If that happens, we know from the balanced equation that for every two moles of octane consumed, we get 18 moles of H2O.

Then we apply that ratio (2/18, or 1/9) to what we put into the reaction (octane). I can invert the ratio from 1 mole octane/9 moles water to (9 moles water/1 mole octane). You'll see why I inverted it in the next step.

Since started with 175 moles of octane, we calculate the moles of water produced as:

(9 moles H2O/1 mole octane)*175 moles octane = 1580 moles H2O (with 3 sigs). [The moles octane cancel out, with the inverted conversion factor). Then convert moles H2O into mass H2O with the molar mass of water (2 H = 2, 1 O = 16, for a total of 12g/moles). [I inverted the conversion so that I could multiply it instead of dividing. It is easier for me to visualize the result with a multiplication., which I like to do to make sure the answer is within expectations, and not a result of a careless error.

Use the units to figure out how to calculate grams of H2O for 1580 moles of H2O. We have moles, and g H2O/mole. So if we multiply them, the moles will cancel, leaving just grams.

1580 moles H20 * (18 g/mole H2O) = 14210 grams. Correct for 3 sig figs to arrive at 28400 grams, or 28.4 kg.

I hope this helps,

Bob

Feel free to ask for clarification.