Question

Two bulbs are connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.700 atm, 0.700 atm, and the small bulb, with a volume of 1.50 L, contains oxygen at a pressure of 2.50 atm. The temperature at the beginning and the end of the experiment is 22 ∘ C 22∘C . After the stopcock is opened, the gases mix and react. 2 NO ( g ) + O 2 ( g ) → 2 NO 2 ( g ) 2NO(g)+O2(g)→2NO2(g)

Which gases are present at the end of the experiment?

Answer 1

`O_(2)` and
`NO_(2)`

Step-by-step explanation:

For a given system at constant temperature, the number of moles of gas present in the system is proportional to the product of the system pressure and volume. Therefore, we have:

NO: 6 L * 0.7 atm = 4.2 L*atm

O: 1.5 L* 2.5 atm = 3.75 L*atm

For the given system based on a balanced chemical equation:

2.70 L*atm of nitric oxide reacts with (2.7/2) 1.35 L*atm of oxygen. This shows that there is more oxygen gas in the system than nitric oxide. Thus nitric oxide is the limiting reactant.

At the end of the experiment:

All the nitirc oxide has been used up, i.e.
`P_(NO)` = 0

For the product: 2.70 L*atm NO produced 2.70 L*atm
`NO_(2)`

The total volume of the system after the stopcock is opened = 6+1.5 = 7.5 L

The partial pressure of
`NO_(2)` = (2.70 L*atm
`NO_(2)` ) / (7.5 L) = 0.36 atm
`NO_(2)`

Similarly for oxygen gas:

3.75 L*atm - 1.35 L*atm = 2.40 L*atm oxygen gas remaining

Partial pressure of oxygen is:

2.40 L*atm / 7.5 L = 0.32 atm

Thus, the gases present at the end of the experiment are
`O_(2)` and
`NO_(2)`