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)))# = #((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))#

May 25, 2017

Answer:

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#