Using the following equation, determine the volume of oxygen gas at 320 K and 680 torr which will react with 2.50 L of NO gas at the same temperature and pressure.
2NO(g) + O2(g) -------> 2NO2(g)
a. 1.25 L
b. 2.50 L
c. 3.00 L
d. 1.00 L
e. 5.00 L
The volume of the oxygen at 320 K and 680 torr that will react with 2.5 L of NO at the same conditions is 1.25 L
Explanation:
There is should be assumed that all the gases behave as ideal gases: PV = nRT, where P is the absolute pressure, V is volume, n is mole number, R is the universal constant of gases: 62.364 L torr/mol K, and T is the absolute temperature.
First the number of moles of NO are calculated as: [tex]n_{NO}=\frac{P_{NO}V_{NO}}{RT_{NO}}=\frac{680*2.5}{62.364*320}=0.08518[/tex] mol of NO.
Then, using the  stoichiometric relationship from the balanced reaction equation the number of moles of O2 are calculated: [tex]0.08518*\frac{1}{2} =0.04259[/tex] mol of O2.
Finally, using the ideal gas equation and solving for the O2 volume: [tex]V_{O_2}=\frac{n_{O_2}RT_{O_2}}{P_{O_2}}=\frac{0.04259*62.364*320}{680}=1.25[/tex] L of O2.
