1 mole of a sample of potassium nitrate decomposes in a 1-L container heated to 457.67 ℃. 2KNO3(s) → 2 KNO2(s) + O2(g)

Question

1. This reaction yields 
2.    grams of O₂ gas, which exerts a pressure of 

   atm against the walls of the container.

Answer 1

Since the mole ratio of O2 to KNO3 is 1:2, you will produce .5 mol of O2 gas.

1 mol KNO3. x 1 mol O2/2 mol KNO3 = 0.50 mol O2

We were given the temperature of O2 and the volume of O2 in the question so using the moles of O2 we just found, we can use the ideal gas law (pv= nRT) to solve for the pressure.

First convert celcius temp to kelvin:

K= 457.67 C. + 273= 730.67 K

now solve for pressure. P=nRT/v:

p=(.5 mol)(8.314 KPa* L/mol*K)(730.76 K)/(1L)= 3037 kPa