Question

Partial Pressure/Total Pressure Question

Consider the equation:

(CH3)2N2H2(l) + 2N2O4(l) ---> 3N2(g) + 4H2O(g) + 2CO2(g)

150.0g of (CH3)2N2H2(l) reacts with excess dinitrogen tetroxide and the product gases are collected at 127°C in an evacuated 250L tank. Assuming the reaction has a 100% yield and all gases behave as ideal gas.

A) How many moles of N2 could be produced after the reaction completed?

B) What is the partial pressure of nitrogen gas produced?

C) What is the total gas pressure in the tank after the reaction?

Answer 1

Reaction:

(CH3)2N2H2(l) + 2N2O4(l) ---> 3N2(g) + 4H2O(g) + 2CO2(g)

Mass of (CH3)2N2H2 = 150.0 g

T = 127 deg C = 127 deg C + 273.15

=400.15 K

Volume of tank = 250 L

A). Calculation of moles of N2

Moles of N2

= Moles of (CH3)2N2H2(l) x 3 mol N2 / 1 mol (CH3)2N2H2(l)

=(150.0g /60.098 g / mol) x 3 mol N2 / 1 mol (CH3)2N2H2(l)

=2.50 mol N2

B)

Calculation of partial pressure of N2

pV = nRT

p = pressure in atm

V = volume in L

n = number of moles

R = gas constant = 0.08206 L atm (Kmol)^-1

T = 400.15 K

Lets plug all the value and get the partial pressure of N2

p = nRT / V

= 2.50 mol x 0.08206 L atm (Kmol)^-1 x 400.15 K / 250 L

=0.328 atm

Partial pressure of N2 = 0.328 atm

C).

After the reaction there are three gases left :

N2 , H2O and CO2

Calculation of moles of H2O

= 2.50 moles of N2 x 4 mol H2O / 3 mol N2

= 3.33 mol H2O

Calculation of moles of CO2

= 2.50 mol N2 x 2 mol CO2 / 3 mol N2

=1.66 mols CO2

Total moles = (2.50+3.33+1.66) mol

Total pressure = n RT / V

= 7.49 mol x 0.08206 L atm (Kmol)^-1 x 400.15 K / 250 L

= 0.983 atm

Total pressure of tank = 0.983 atm