At STP, what is the mass of 11.2 liters of O2 gas?

Question

I tried solving this myself and got 16.0 g, but wasn't quiet to sure if I did the problem right.

Answer 1

While the previous apprach could have been used, when the system is at STP, you dont have to use PV = nRT. All you need to know is at STP 1 mole of an ideal gas = 22.4 liters. Thus, 11.2 L x 1 mol/22.4 L = 0.5 moles

0.5 moles x 32 g/mole = 16 g.  

You were correct. 

The previous answer has several errors. First , n = 0.499 =0.5 mokes, not 0.0446. Second error is that molar mass of O2 is 32 g, not 16 g. 

Answer 2

You need the equation: PV=nRT

 

P=1atm

T=273K

V=11.2L

R=0.0821

n=?

 

rearrange the equation in terms of n:

 

PV=nRT - divide both sides by RT

 

n=PV/RT

 

now put in the info:

1x 11.2/ 273x0.0821

11.2/22.4133

=0.0446 moles of O2

 

now find the mass by:

0.0446 moles x atomic mass of O2

=0.0446 x 15.99(2)= 1.426g (3 sig fig)

 

you have 1.426g of O2