Question

The total pressure of a mixture of oxygen and hydrogen is 1.00 atm. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.400 atm when measured at the same values of T and V as the original mixture. What was the composition of the original mixture in mole percent?

Answer 1

Answer:

molar composition of oxygen =0.2 (20%)

molar composition of hydrogen = 0.8 (20%)

Explanation:

Since hydrogen and oxygen react according to

2*H₂+ O₂ → 2*H₂O

according to the ideal gas law:

P₁*V=n₁*R*T (initial state)

P₂*V=n₂*R*T (final state)

dividing both equations

P₁/P₂ = n₁/n₂

then since 2 moles of hydrogen react for every mole of oxygen that reacts, thus oxygen that reacted "no" is

no =  (n₁-n₂) 1/(1+2) = n₁ (1-n₂/n₁)/3

then

molar composition of oxygen = xo= no/n₁ = (1-n₂/n₁)/3 = (1-P₂/P₁)/3 = (1-0.4 atm/1 atm )/3 = 0.2 (20%)

then

molar composition of hydrogen = xh= 1- xo = 0.8 (20%)

to verify it , the number of initial moles n₁=remaining hydrogen + hydrogen that reacted + oxygen that reacted

and since 2 moles of hydrogen react for every mole of oxygen that reacts n₁= n₂ + (n₁-n₂) (2/3) + (n₁-n₂) (1/3) = n₂ + (n₁-n₂) = n₁

Answer 2

Answer:

oxygen = 0.2 atm

hydrogen = 0.8 atm

Explanation:

the equation of the reaction:

$2H_{2} + O_{2}$ → $2H_{2}O$

pressure ∝ number of moles

total number of moles of reactant = 3

therefore number of moles of each reactant,

$nO_{2}$ = $\frac{1}{3} O_{2}$  n=1/3

$nH_{2}$ = $\frac{2}{3} H_{2}$  n=2/3

After ignition, the pressure is 0.4 atm, pressure befofre ignition is therefore 1 atm - 0.4 atm = 0.6 atm

$nO_{2}$ = $\frac{1}{3} * 0.6 atm$ = 0.2 atm

$nH_{2}$ = $\frac{2}{3} * 0.6 atm$ = 0.4 atm

Therefore, pressure of $O_{2}$ = 0.2 atm

total pressure of $H_{2}$ = pressure before ignition + pressure after ignition

                                 = 0.4 atm + 0.4 atm

                                 = 0.8 atm