mols O3 = g/molar mass = 4.8/48 = 0.1
Pressure before O3 decomposes is then
P = nRT/V = ?
After decomposition you have
2O3 ==> 3O2
0.1 mol O3 x (3 mols O2/2 mols O3) = ? and use P = nRT/V again to find the O2 pressure.
Calculate the pressure of 4.80g of ozone, O3, in a 2.45 L flask at 25 degrees C. Assume that the ozone completely decomposes to the molecular oxygen.
2O3(g) --> 3O2(g)
Calculate the pressure inside the flask once the reaction is complete.
4 answers
ok I set up the first one right, and got P=.998 atm
The after composition makes sense and I got P = 20.0 atm
Are these correct?
The after composition makes sense and I got P = 20.0 atm
Are these correct?
0.998 is ok. 20.0 is not.
0.1 mol x (3 mols O2/2 mols O3) = 0.1*1.5 = 0.15 and if you substitute 0.15 for n in the equation surely you don't get 20 atm. Wouldn't the correct answer logically be P2 = 1.5*P1 = 0.998*1.5 = about 1.5?
0.1 mol x (3 mols O2/2 mols O3) = 0.1*1.5 = 0.15 and if you substitute 0.15 for n in the equation surely you don't get 20 atm. Wouldn't the correct answer logically be P2 = 1.5*P1 = 0.998*1.5 = about 1.5?
Thanks!