Question

How many liters of NO2 gas at 21C and 2.31 atm must be consumed to produce 75.0 L of NO gas at 38C and 645 torr according to the following balance reaction?

3NO2 + H2O ----> 2HNO3 + NO

Answer 1

PV = nRT

where, P = pressure = 645 torr = 0.849 atm

V = volume = 75.0 L

n = number of moles

R = Gas constant

T = temperature = 38 + 273 = 311 K

0.849 * 75 = n * 0.0821 * 311

6.16 = n * 25.5

n = 6.16 / 25.5 = 0.242 mole of NO

From the balanced equation we can say that

1 mole of NO is produced by 3 mole of NO2 so

0.242 mole of NO will be produced by 0.726 mole of NO2

PV = nRT

2.31 * V = 0.726 * 0.0821*294

2.31 V = 17.5

V = 17.5 / 2.31 = 7.58 L

Therefore, the volume of NO2 consumed would be 7.58 L