posted by JRMM on .
Molar volume of an ideal gas at S.A.T.P. (25oC and 1.00 bar is 24.8 dm3 mol-1. Assume that gases are at 225oC and P=1.00 atmosphere before and after the combustion and behave as ideal gases.
The chemical equation for the combustion of butane is:
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)
At 225oC, we ignite a mixture of 5.00 dm3 of butane and 75.0 dm3 of O2
i. Is there enough oxygen for the complete combustion of the butane present?
ii. What gases are present in the resulting mixture after combustion?
iii. Calculate the final volume of the mixture after combustion (3 sig. fig).
Iv. Calculate the amounts (number of moles) of O2 and butane consumed using the molar volume of an ideal gas at S.A.T.P (3 sig. fig).
v.The final mixture is cooled down from 225oC to 25oC. Calculate the volume of the resulting gaseous mixture (3 sig. fig.).
There is a trick on using SATP. Note that one bar is not one atmosphere, If I recall, it is .986atm. Check that.
Secondly, the degree symbol (little superscripted zero) is not used with C or K, you just write 25C, or 300K, and when read, you do NOT say degrees. Twenty-five celcius is correct.
The problem is when you write 225oC, that is either 225C ,or 2250C, which are plenty different.
Now on your problem, calculate the moles of butane and O2 first. Look at the mole ratio in the balanced equation, which is deficient in quantity? After the reaction, you have along with the products, the reactant which was in excess?