How many moles of methanol must react with excess oxygen to produce 5.0 L of carbon dioxide at STP?

1 Answer
Jul 25, 2017

Answer:

0.223 moles

Explanation:

The equation for the combustion of methanol is:

#2CH_3OH(l) + 3O_2(g) → 2 CO_2(g) + 4 H_2O(g)#

Firstly you need to work out how many moles of carbon dioxide are in 5 litres at STP. If you assume that it is an idea gas (it isn't, but it's close enough for this sort of purpose) then 1 mole at STP occupies 22.4 litres. Therefore, at STP 5 litres is #(5/22.4) = 0.223# mol.

Looking at the equation, it tells you that 2 moles of carbon dioxide are produced by complete combustion of 2 moles of methanol. Therefore, if you have 0.223 moles of carbon dioxide, you would need 0.223 moles of methanol.

Molar mass of methanol is 32.04 g/mol so you would need 0.223 x 32.04 = 7.14 g of methanol.