What volume of O2 (m3) is needed for the complete combustion of 155 g of C3H8 at 19.0oC and?

Question

What volume of O2 (m3) is needed for the complete combustion of 155 g of C3H8 at 19.0oC and 100.0 kPa

Answer 1

Time to do some equation solving:
C3H8 + 5O2 -> 3 CO2 + 4 H2O
So now its time for the three step drill
Find the mol wt of propane (like 44)
Convert the mass of propane to moles (like 155/44)
Find the number of moles of oxygen needed to carry out the equation (like 5 x 155/44)
Once you have that number, its the pV= nRT routine where p = 100 kPa and T= 292 K and n is the number of moles from above. NO SWEAT

Answer 2

C3H8+n*O2 --> m*CO2+p*H2O m should be 3 because you've 3 carbons on the left end aspect, so that you eventually end up with 6+p form of oxigen atoms. p should be 4 because you've 8 H on the left hand aspect, which signifies that n=6+p or n=10. So, for each mole of C3H8 you burn you opt for 10 moles of O2. next step is to calculate what number moles you've in one hundred fifty 5 g of C3H8, multiply that by technique of 10 and discover the volume occupied by technique of a lot oxigen on the temperature and preassure you specify.