Respuesta :
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:
oxygen = 0.2 atm
hydrogen = 0.8 atm
Explanation:
the equation of the reaction:
[tex]2H_{2} + O_{2}[/tex] → [tex]2H_{2}O[/tex]
pressure ∝ number of moles
total number of moles of reactant = 3
therefore number of moles of each reactant,
[tex]nO_{2}[/tex] = [tex]\frac{1}{3} O_{2}[/tex] n=1/3
[tex]nH_{2}[/tex] = [tex]\frac{2}{3} H_{2}[/tex] n=2/3
After ignition, the pressure is 0.4 atm, pressure befofre ignition is therefore 1 atm - 0.4 atm = 0.6 atm
[tex]nO_{2}[/tex] = [tex]\frac{1}{3} * 0.6 atm[/tex] = 0.2 atm
[tex]nH_{2}[/tex] = [tex]\frac{2}{3} * 0.6 atm[/tex] = 0.4 atm
Therefore, pressure of [tex]O_{2}[/tex] = 0.2 atm
total pressure of [tex]H_{2}[/tex] = pressure before ignition + pressure after ignition
= 0.4 atm + 0.4 atm
= 0.8 atm