Question

The world burns approximately 3.4×1012 kg of fossil fuel per year. Use the combustion of octane as the representative reaction and determine the mass of carbon dioxide (the most significant greenhouse gas) formed per year.

Answer 1

@abb0t

Answer 2

So firstly what we will need to do is get an equation so we know what we are working with. The equation we need is the combustion of octane which will be this when its balanced:

2C8H18 + 25O2 ----> 16CO2 + 18H2O

So firstly, you have the amount of fossil fuel in your question which I'm guessing would be octane. So we know we have 3.4×10^12 kg of octane in our equation.

We need to find the moles of octane so we can use it to find the moles of CO2. So we can find the moles of a substance using this equation: moles = mass (g) / molar mass (g mol-1)
Our answer is in kg so I think we should convert that into grams (1kg = 1000g) so 3.4×10^12 kg x 1000 = 3.4x10^15 g of octane
We also have to work out the molar mass: (12x8) + 18 = 144g mol-1

Now we can use the equation:
3.4x10^15g / 144g mol-1 = 2.36 x 10^13 mol of octane (3 sig.fig)

Now we have the moles of octane, we can work out the moles of CO2. As there is 2 moles of octane, and 16 moles of CO2, we have to times our answer of moles by 8 (because the answer we have is for 2 moles, so we have to make that into 16 moles 16/2 = 8)

2.36x10^13 mol x 8 = 1.89x10^14 mol of CO2 (3.sig.fig)

Now we have the moles of CO2, we can convert that into the mass of CO2. In your question it dosent state what units you need the gas in. So, usually gas is in dm3 so I will convert it into that using this: moles x 24

1.89 x 10^14 x 24 = 4.54 x 10^15 dm3 of CO2 gas is created a year.
If you need that answer in cm3, then just times it by 1000.
If you need the answer in L, then times the moles by 22.4L
If you need the answer in g, then times the moles by its molar mass (44)

By the way, all my answers have been assuming that this is all at RTP (room temp and pressure) so if you needed any specific temperatures/pressures etc. hopefully someone else will be able to help you with that.

Either way, I hope this helped even a little! Or even to get you on the right track. Feel free to correct me :)