given that 4g of methane were burnt in 7dm cube oxygenen what is the limiteng reagent?

Question

given that 4g of methane were burnt in 7dm cube oxigen what is the limiting reagent? if 5 dm cube of co2 were produced whatn is the yield of the reaction?

Answer 1

The 4g of methane is a quarter of a mole. ( 1 mole of methane is 16g). A mole of ANY gas occupies 22.4 litres at normal temperature and pressure. So a quarter of a mole has a volume of 5.6 litres.
The equation for the reaction is:
CH4 + 2O2 = CO2 + 2H2O
This shows that twice as much oxygen is needed as methane.
So to burn 5.6 litres of methane you need 11.2 litres of oxygen.
This means that the limiting reagent is oxygen as there isn't enough.
If 5dm3 of CO2 is made then more oxygen must have been added as you would only expect to get the half the volume of CO2 than the volume of oxygen you started with. In your case you started with 7 dm3 of oxygen and so the maximum amount of CO2 is 3.5 litres.
Assuming the 4g of methane were burned in EXCESS oxygen you would expect to get 5.6 litres of CO2 and in this case the % yeild is 5 divided by 5.6 multiplied by 100 = 89.3%

At room temperature the figures would be slightly different. i.e 6 dm3 of CO2 would be expected and the yeild would be 83.33%

So you can see that it is important to say at what temperatures the gases were measured as they tend to expand when they are warmed.

Answer 2

We will assume room temperature and pressure. 4g of methane is 1/4 of a mole, ie 6dm3 (at room temperature and pressure). This will require 12dm3 of oxygen gas for complete combustion. So thjere isn't enough oxygen.
If 5dm3 of CO2 is produced, then 5dm3 of methane was used up, leaving 1dm3 over. So the yield is 5/6. But the pressure and temperature need to be stated, and, perhaps, that figure of 24 dm3 for the molar volume changed.

Answer 3

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