Answer:
Mass of O₂ present in the room: 185200 g
185.2 kg
Quantity of O₂ = 5787.5
Explanation:
The ideal gases Law can solved, what is ths mass of O₂ present in the room, or the quantity of O₂ (moles), present in the room.
P . V = n . R . T
P = 1 atm
V = volume of the room
n = moles
R = 0.082 L.atm/mol.K
T = T° in K
T° C + 273 → 22°C + 273 = 295K
Let's find out the volume of the room
7 m . 8 m . 2.5 m = 140 m³
Units of R, are in L, so we must convert m³ to dm³
1dm³ = 1 L
140m³ . 1000 = 140000 dm³
1.4x10⁵L
1 atm . 1.4x10⁵L = n . 0.082 L.atm/mol.K . 295K
(1 atm . 1.4x10⁵L) / (0.082 L.atm/mol.K . 295K) = n
5787.5 moles = n
molar mass . moles = mass
5787.5 moles O₂ . 32 g/m = 185200 g of O₂
