Question

Question #56fe1

Answer 1

One mole any gas at STP => 22.4 Liters
Given moles of `O_2(g) = "mass(g)/(F.Wt(g/(mol)))` = `((65g)/(32(g/(mol))))` `2.03(mol O_2(g))`

Volume of 2.03 mole `O_2(g)` at STP = `(2.03mol)(22.4(L/(mol))`
= `45.47(LitersO_2(g))`

Answer 2

The volume is 45.5L.

Explanation:

Molecular mass of oxygen is 32g.
32g `O_2` has a volume of 22.4L
Hence,the volume of 65g oxygen in STP is `22.4/32×65=45.5L`