Gas Law Problem 120 mL of NH3 at 25*C and 750 torr was mixed with 165 mL of O2 at 50*C and 635 torr and transferred to a 300 mL reaction vessel where they were allowed to react accoring to the...
Gas Law Problem
120 mL of NH3 at 25*C and 750 torr was mixed with 165 mL of O2 at 50*C and 635 torr and transferred to a 300 mL reaction vessel where they were allowed to react accoring to the equation:
4NH3 (g) + 5O2 (g) -> 4NO(g) + 6H20 (g).
What was the total pressure (in torr) in the reaction vessel at 150*C after the reaction is over? Assume the reaction goes to completion.
Start with how many mL of each reactant you have at STP, using combined gas law.
NH3: 750*120/298 = 760*V2/273 = 108.49 mL
O2: 165*635/323 = 760*v2/273 = 116.52 mL
Convert volume to moles, knowing there are 22400 mL /mole.
NH3: 108.49/22400 = .00484 moles
O2: 116.52/22400 = .0052 moles.
Now look at your mole ratios in the balanced equation. For every 4 moles of NH3 you need 5 moles of O2, and vice-versa.
If you have .00484 moles of NH3, you need 5/4*.00484 moles of O2 = .00605 moles of O2 but you only have .0052 moles. So the O2 is the limiting reactant.
The 0.0052 moles of O2 will combine with 0.00416 moles of NH3, leaving .00068 moles unreacted.
The 0.0052 moles of O2 will produce 0.00416 moles of NO + .00624 moles of HOH.
So in the .3 L reaction vessel you have 0.00068 moles of NH3 + 0.00416 moles of NO + 0.00624 moles of HOH, or a total of 0.01108 moles.
Now use the universal gas law: PV = nRT
you know V = 0.3 L, n = 0.01108 moles, R = 8314.51 and T = 423 K.
P = 129,896 Pa or 129.9 kPa