What volume of oxygen gas at 25°C and 1.28 atm is needed for the complete combustion of 5.40 g of propane?

I am assuming that that is a general chemistry question and so there are some assumptions to be made. That we are treating both propane and oxygen as ideal gases.

first you need to balance your equation it should look like this so that you have even moles on each side.

C3H8 (g) + 5 O2(g) ---> 3CO2(g) + 4 H2O (g)

so looking at the equation we know that there are 5 moles of oxygen consumed per each mole of propane.

with this knowledge we find the moles of propane being used by dividing by the molar mass of 44.1 g/mol. 5.4 g/ 44.1 g/mol = 1.22449 mole propane

multiply by 5 to find the number of moles of oxygen used. 0.61224 mol O2

We then have to used PV=nRT in order to solve. (This is the point at which it being and ideal gas is very important, if it is not you have to bring in activity coefficients, let me know if that is the case.)

Plug everything you have in for what you know treating temp. in K.

(1.28atm) (V) = (.61224 mol O2) (0.0821 L atm/ K mol) (298 K)

solving for V you get 11.93 L. Remember to watch sigfigs, here I think the exact answer should be 11.9 L.

CH4+2O2=CO2+2H2O MOL RATIO=1:2 VOLUME OF OXYGEN=2*2.36=4.72 D

Mr of propane= 12x3 + 8 = 44

mole of propane used= 5.4/44= 27/220 mole

1 mole of propane uses 1 mole of oxygen, therefore,

27/220 mole of oxygen is used.

using the formula pV= nrT,

V= nrT/p

=(27/220)(8.315)(25+273)/(1.28x101.03)

=2.35 dm3

n(C3H8) = m/M

m = 5.40g

M = (3 x 12) + (1 x 8) = 44

therefore

n(C3H8) = 5.40/44

n(C3H8) = 0.123

n(C3H8) = n(O2)

therefore

n(O2) = 0.123

General gas equation:

pV = nRT

V = (nRT)/p

n = 0.123

R = 8.31 (constant)

T = 25C = 298K

p = 1.28 atm = 129.696kPa

therefore

V(O2) = (0.123 x 8.31 x 298)/129.696

V(O2) = 2.349L