# Stoichiometric calculations for the combustion of octane [closed]

The equation for the burning of octane:

$$\ce{2C8H18 + 25O2 -> 16CO2 + 18H2O}$$

1. How do i find the number of moles of carbon dioxide which is produced when one mole of octane burns ?
2. How do i find , at what volume , at STP , is occupied by the number of moles determined in the answer to my first question ?
3. If the relative molecular mass of carbon dioxide is 44 , what is the mass of carbon dioxide produced by burning 2 moles of octane?

## 1 Answer

This is a homework problem, so I'll not spell the answer outright, but try and give as much as I would if you came to me in office hours.

1. Your equation tells you what the ratio of moles of $\ce{C8H18}$ to moles of $\ce{CO2}$. For every two moles of octane burned, how many moles of $\ce{CO2}$ are produced? Now, what would it be for only one mole of octane?

2. What have you tried so far on this? You have the ideal gas law (PV = nRT), and you know the P, the n from the first question, and the T - so just solve for the V.

3. So you know one mole of $\ce{CO2}$ masses 44 grams, so just multiply this by the number of moles of $\ce{CO2}$ produced by 2 moles of octane (see my hint on point 1 above).