[H Mac]

The question being...

Calculate the mass of potassium chlorate needed to prepare 5.00 L of oxygen gas at 24.0 'C and 0.950 atm. (When using atmospheres, the gas constant, R = 0.0821 L/atm//mol -7/K-1).
The equation is KClO3 -->  KCl + 3/2 O2.

Therefore:
P= 0.950 atm
V= 5.00 L
n=  PV
     RT
R= 0.0821 L/atm/mol-7/K-1
T= 297K

Therefore the number of moles in O2 is...
n =         (0.950 atm * 5.00 L)
     ( 0.0821 L/atm/mol-7/K-1 * 297 K)
   = 0.195 mol-7

Therefore make your ratios
    x mol              0.195 mol
    KClO3                 O2
    1 mol               1.5 mol

     x mol   =   0.195 mol
     1 mol         1.5 mol
 
     x mol = 0.13 mol
The mass of KClO3 = 122.6 g/mol
Therefore  0.13 mol * 122.6 g/mol = 15.9 g

The mass of potassium chlorate needed is 15.9 g

Does this look correct to you?

[Borek]

Seems OK

[H Mac]

Thank you for this help... Its helping move on with my chemistry much faster than I ever thought I would be able to!!! Thank you again!

[AWK]

But what does mean this unit at R

R= 0.0821 L/atm/mol-7/K-1