What volume of O2 (m3) is needed for the complete combustion of:?

Question

138 g of C3H8 at 25.0oC and 100.0 kPa?

Answer 1

A balanced equation for this combustion is:

C3H8 + 5 O2 ----> 3 CO2 + 4 H2O

So, every mole of C3H8 requires 5 moles of O2 to burn.

# of moles of C3H8 = 138 / (12*3+8) =3.136 mole

# of moles of O2 required = 3.136*5 = 15.682 moles

Using ideal gas law, PV = nRT, we get

V = nRT / P = 15.682*8.314*(25+273) / (100*1000)
V = 0.389 m^3

Answer 2

Convert 138 g C3H8 to moles by dividing by the molar mass (44g/mole) = 3.14 moles

Write and balance the equation:

C3H8 + 5 O2 = 3CO2 + 4 H20

From the balanced equation it can be seen the mole ratio between C3H8 and O2 is 1:5 (look at the coefficients in the balanced equation).

5 times as many moles of oxygen are produced so multiply the moles of C3H8 (3.14) by 5 = 15.7 moles of oxygen.

3.14 moles C3H8 x 5 moles O2/mole C3H8 = 15.7 moles O2

Avogardro's hypothesis states 1 mole of a gas occupies 22.4 liters at STP (0 degrees C and 101.325 kPa). Multipy the moels of oxygen produced in the reation (15.7) by 22.4.

15.7 moles x 22.4L/mole = 351Liters at STP

Use the Combined gas law to change the volume at STP to the volume at 25 degrees and 100.0 kPa.

P1 = 101.325 kPa P2 = 100.0 kPa
V1 = 351 Liters V2 = x
T1 = 0 + 273 = 273 Kelvin
P2 = 100.0 kPa
V2 = x
T2 = 25 + 273 = 298 Kelvin

(101.325 )(351)/273 = (100.0)(X)/298

x = 388 Liters O2 @ 25 C and 100.0 kPa