Ethane burns in air to give H 2 O and CO 2 . 2 C 2 H 6 (g) + 7 O 2 (g) 4 CO 2 (g

Question

Ethane burns in air to give H2O and CO2.
2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)
(a) Four gases are involved in this reaction. Place them in order of increasing rms speed. (Assume all are at the same temperature.)
(b) A 3.26-L flask contains C2H6 at a pressure of 256 mm Hg and a temperature of 25 °C. Suppose O2 gas is added to the flask until C2H6 and O2 are in the correct stoichiometric ratio for the combustion reaction. At this point, what is the partial pressure of O2 and what is the total pressure in the flask?

Explanation / Answer

a)

RMS speed = sqrt(3*RT/MW)

the higher the molar weight, the lower the velocity, therefore choose the lightest first

MW C2H6 = 30

MW O2 = 32

MW CO2 = 44

MW H2O = 18

Velocity (RMS)

H2O > C2H6 > O2 > CO2

b)

V = 3.26 L

P = 256 mm Hg = 0.336 atm

T = 25°C = 298 K

if you add O2 util they are stoichiometri

find P O2 in the pressure of the flask

First, calculate mol of gas

n = PV/(RT) = 0.336*3.26/(0.082*298) = 0.044825 mol

N1 = 0.044825 P1 = 256

For the second case

mol of O2 added

due to stoichiometry = 0.044825 mol of C2H6 need 7/2*0.044825 mol O" = 0.1568 mol of O"

total mol = 0.1568+0.044825 = 0.201625 mol

Then

P1/n1 = P2/n2

P2 = P1*(n2/n1) = 256*(0.201625/0.044825) = 1151.500 mm HG

the mole fraction of O2

mol frac O2 = mol O2 / total mol = 0.1568/0.201625 = 0.77768

then

P O2 = mol frac O2 * PT = 0.77768*1151.500 = 895.498 mm Hg of O2

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