The picture below shows two bulbs connected by a stopcock. The large bulb, with

Question

The picture below shows two bulbs connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.400 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 degree C. After the stopcock is opened, the gases mix and react: Which gases are present at the end of the experiment? NO O_2 NO_2 What are the partial pressures of the gases? If the gas was consumed completely, put 0 for the answer. P_NO = P_O_2 P_NO_2 =

Explanation / Answer

2NO(g) + O2(g) --> 2NO2(g)

PV = nRT
n = PV / RT
n = 0.40 atm * 6.00L / 0.0821 Latm/molK / 295K = 0.099 mol NO
n = 2.50 atm * 1.50L / 0.0821 Latm/molK / 295K = 0.155 mol O2

1 mol of O2 requires 2 moles of NO, therefore, 0.155 mol O2 requires 0.310 moles of NO, but there are only 0.099 moles of NO.Therefore, NO is the limiting reactant. All of the NO is consumed.

Therefore, 0.099 moles of NO2 is produced, and 0.0495 moles of O2 reacts, leaving an excess of 0.1055 moles of O2.

Partial pressure of NO2 = nRT / V
P = 0.099 mol x 0.0821 Latm/molK x 295K / 7.50L
P = 0.320 atm NO2

P = 0.1055 mol x 0.0821 Latm/molK x 295K / 7.50L
P = 0.34 atm O2

Of course the partial pressure of NO is zero.

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