Respuesta :
Answer:
mass O2 = 3,456 10⁻² gr
Explanation:
To solve this problem we use the equation of state of ideal gases
PV = nRT
With the initial data we calculate the initial amount of gas
n = PV / RT
Let's reduce the magnitudes to SI units
P1 = 755.0 torr (1 10⁵Pa / 760 torr) = 0.99 10⁵ Pa
V1 = 2.20 L (m3 / 10³ L) = 2.20 10⁻³ m3
P2 = 710.2 = 0.93 10⁵ Pa
n = 0.99 105 2.20 10⁻³ / (8.314 292)
n = 8.97 10⁻² mol
Now we can calculate the moles consumed
P1 V = n1 RT
P2 V = n2 RT
P1 / P2 = n1 / n2
n2 = n1 P2/P1
n2 = 8.97 10⁻² 0.93 10⁵ / 0.99 10⁵
n2 = 8.43 10⁻² mol
These are the moles of air left in the chamber, the moles consumed is the difference with the initials
Δn = n1-n2
Δn = 8.97 10⁻² - 8.43 10⁻²
Δn = 0.54 10⁻² mol = 5.4 10⁻³ mol
These are the moles of air consumed, in the normal atmosphere only 20% is oxygen, therefore, the mole of oxygen is
n O2 = 0.20 5.4 10⁻³
n O2 = 1.08 10⁻³ mol
The oxygen molecule has two atoms and the molecular weight of oxygen is 15,999 gr / mol
PM = 2 PA = 32 gr / mol
mass O2 = Pm mol
mass O2 = 32 1.08 10⁻³
mass O2 = 3,456 10⁻² gr
