Question

2. Ammonia combines with oxygen gas to form nitrogen gas and water vapor. A 3.55-L container of ammonia kept at 280 K and 1.66 bar is reacted with a 8.19 L of oxygen measured at 302 K and 0.776 bar. What is the volume of the final gas mixture measured at STP (1 bar, 273 K)?

Answer 1

2)

volume of NH3 = 3.55 L

temperature = 280 K

pressure = 1.66 bar

P V = n R T

1.66 x 3.55 = n x 0.0821 x 280

n = 0.2564

moles of NH3 = 0.2564

P V =n R T

0.776 x 8.19 = n x 0.0821 x 302

n = 0.2563

moles of O2 = 0.256

4 NH3   + 3 O2    ---------------> 2 N2 +   6 H2O

   4               3                                2             6

0.256          0.256

here limiting reagent is NH3 .

moles of N2 fomed = 0.256 x 2 / 4 = 0.128

moles of H2O = 0.384

moles of O2 remain = 0.064 mol

total moles = 0.576

P V = n R T

1 x V = 0.576 x 0.0821 x 273

V = 12.9 L

final volume = 12.91 L