Nitric acid, a very important industrial chemical, is made by dissolving the gas nitrogen dioxide (NO2) in water. Calculate the moles of gas produced in 2 L volume container, 736 torr and 50 ºC if the gas behaves as ideal:

I haven't done this in years... but looks like you can do this using the formula PV = nRT if I remember correctly where P = pressure in atm, V = volume in L, n = number of moles, R = 0.0821 (atm*L*K)/mol "I'm not so sure about the unit on R", and T = temperature in Kelvin.

NO2 is not the anhydrous form of HNO3. N2O5 is. When NO2 is dissolved in water a mixture of HNO2 and HNO3 is produced. Commercially, O2 is added to oxidize the HNO2 to HNO3. Since the HNO3 produced is an aqueous solution, I don't understand calculating mols of a gas.

Concerning R, it the units are L*atm/mol*K.
2NO2 + H2O ==> HNO2 + HNO3

To calculate the moles of gas produced in the given conditions, we will use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = moles of gas
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)

First, let's convert the given pressure from torr to atm, and the given temperature from Celsius to Kelvin.

736 torr * (1 atm / 760 torr) = 0.9684 atm
50 ºC + 273.15 = 323.15 K

Now, we can substitute the values into the ideal gas law equation:

PV = nRT
(0.9684 atm) * (2 L) = n * (0.0821 L·atm/(mol·K)) * (323.15 K)

Simplifying the equation:

1.9368 = n * 26.530615

Dividing both sides by 26.530615:

n = 1.9368 / 26.530615 ≈ 0.073 mol

Therefore, approximately 0.073 moles of gas are produced in a 2 L volume container at 736 torr and 50 ºC, assuming the gas behaves ideally.