[simonp]

Hi,

I have to answer some mass and ratio questions on the combustion of propane

a) combustion equation - done
b) mass of , oxygen carbon dioxide and water vapour for each 1g of fuel consumed - done
c) mass of air for complete combustion of 1g of fuel.  

I have worked the combustion equation out

Propane:
C3H8    +     5O2          =        3CO2    +    4H2O
1 mole       5 moles        3 moles      4 moles
44g        160g             132g      72g   mass
1g        3.64g            36.26g      19.78g    ratio

For the mass of air question I know that oxygen is 21% and Nitrogen 79% (ignoring other minor gases).  First I thought that the mass of air would simply be 5 x 3.64g, but that would not be precise.  Then I wondered whether the 21%/79% are volumes so I would need to convert somehow using V=m/d.  Now I'm just confused. 

[Borek]

21%/79% are volume percentages - which means (according to Avogardo's law) that for every 21 moles of oxygen there is 79 moles of nitrogen.

[AWK]

Chemical engineers use O₂+4N₂ for air content.
This gives about 2% error for mass or volume of air.