Question #56fe1

2 Answers
May 25, 2017

One mole any gas at STP => 22.4 Liters
Given moles of O_2(g) = "mass(g)/(F.Wt(g/(mol)))O2(g)=massgF.Wt(gmol) = ((65g)/(32(g/(mol))))65g32(gmol) 2.03(mol O_2(g))2.03(molO2(g))

Volume of 2.03 mole O_2(g)O2(g) at STP = (2.03mol)(22.4(L/(mol))(2.03mol)(22.4(Lmol)
= 45.47(LitersO_2(g))45.47(LitersO2(g))

May 25, 2017

The volume is 45.5L.

Explanation:

Molecular mass of oxygen is 32g.
32g O_2O2 has a volume of 22.4L
Hence,the volume of 65g oxygen in STP is 22.4/32×65=45.5L22.432×65=45.5L