Question

a cylinder of volume 1.22*10^-3 m^3 contains methane at 15 degrees centigrade. the pressure gauge on the cylinder shows the methane to be at a pressure of 3.28*10^5 Pa. what volume of O2 at 25 degrees centigrade and a pressure of 1*10^5 Pa will be required to completely burn all the methane? gas constant R=8.314JK^-1 mol^-1

Answer 1

PV=nRT
P=3.28*10^5 Pa
V=1.22*10^-3 m^3
R=8.314JK^-1 mol^-1
T=15 degrees centigrade=(273+15)K=288K
n=PV/RT=(3.28*10^5*1.22*10^-3)/(8.314*288)
n=0.167moles
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
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1mole of methane combines with 2 moles of O2
0.167 of methane combines with x moles of O2
1/2=0.167/x
x=0.334moles of O2
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P(oxy)=1*10^5
V(oxy)=?
n=0.334moles
T=25 degrees centigrade=(273+25)K=298K
V(oxy)=nRT/P(oxy)
V(oxy)=(0.334*8.314*298)/1*10^5 =0.00827509048=
=8.275*10^-3cubic meters.

volume of O2 required to completely burn
all the methane=8.275*10^-3cubic meters.

Answer 2

Thank you!!