This is a limiting reagent (LR) problem. I work these the long way but I think it's easier to explain this way.
C + O2 ==> CO2
mols C = grams/molar mass = 14/12 = approx 1.1 but you need to redo this for a more accurate number.
mols O2 = 58/32 = 1.8
Convert mols C to mols CO2. That's
approx 1.1 x (1 mol CO2/1 mol CO) = approx 1.1
Convert mols O2 to mols CO2. That's approx 1.8 x (1 mol CO2/1 mol O2) = 1.8
In LR problems the smaller amout is ALWAYS the correct answer and the reagent responsible for this is the LR. So C is the LR, O2 is the excess reagent. You want to know volume of gas. Some of that gas will be CO2 produced by the reaction BUT there will be some O2 not reacted and that will contribute to the total volume.
So how much O2 do we have that didn't react?
1.1 mols C x (1 mol O2/1 mol C) = 1.1 mols O2 used.
mols O2 not reacted is 1.8-1.1 = approx 0.7. So total mols gas is 1.1 from CO2 formed + 0.7 from the unreacted O2 = 1.8 total mols gas.
Now used PV = nRT and convert total mols to volume in liters. Post your work if you get stuck. REMEMBER these numbers I've used are estimates. You must go through and rework the problem with better numbers than I've used.
14 grams of coal undergoes complete combustion in 58 grams of oxygen gas.
What volume of gas (measured at 25 degrees celsius and 100 kPa) is present when the reaction is complete?
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