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?

and

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. = 0

For the product: 2.70 L*atm NO produced  2.70 L*atm

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

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

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 and

2b2t

explanation:

2b2t