Question

The following reaction was performed in a 1.60 L non-leaking container at a constant temperature of 134.0 ℃.

4NO(g)+6H2O(g)→4NH3(g)+5O2(g)

Upon completion of the reaction 0.970 g O2 was formed.

A) What was the pressure of O2 after the reaction was completed?

B) What was the minimum starting pressure of NO?

C) What was the minimum starting pressure of H2O?

D) If stoichiometric amounts of NO and H2O were used and the reaction went to completion, what was the total pressure of the system before the reaction took place?

E) If stoichiometric amounts of NO and H2O were used, what was the total pressure of the system after the reaction went to completion?

Answer 1

a) We calculate the moles of O2:

nO2 = 0.97 g * (1 mol / 16 g) = 0.061 mol

Applying the law of ideal gases and partial pressures we have:

PO2 = nO2 * R * T / V = ​​0.061 * 0.082 * 407 / 1.6 = 1.27 atm

b) We calculate the moles of NO needed to produce that amount of O2:

nNO = 0.061 mol O2 * (4 mol NO / 5 mol O2) = 0.049 mol NO

we apply ideal gas law:

PNO = 0.049 * 0.082 * 407 / 1.6 = 1.022 atm

c) same previous procedure:

nH2O = 0.061 mol O2 * (6 mol H2O / 5 mol O2) = 0.073 mol H2O

PH2O = 0.073 * 0.082 * 407 / 1.6 = 1.523 atm.

d) if they were fed stoichiometrically, initially we have 10 mol of ideal gas and we calculate:

Pt = 10 * 0.082 * 407 / 1.6 = 208.59 atm

e) if they were fed stoichiometrically, finally we have 9 mol of ideal gas and we calculate:

Final Pt = 9 * 0.082 * 407 / 1.6 = 187.73 atm.