Question

A room with dimensions 7.00m×8.00m×2.50m is to be filled with pure oxygen at 22.0∘c and 1.00 atm. The molar mass of oxygen is 32.0 g/mol.

Answer 1

Mass of O₂ present in the room: 185200 g

185.2 kg

Quantity of O₂ = 5787.5

Step-by-step 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