Calculating Volume of Air for Nitrogen Dioxide Dillution

[synergix]

Homework Statement

Nitrogen dioxide reacts with water in the atmosphere to produce acid rain.
4N02(g)+2H20(g)+O2(g)------> 4HNO3(g)
assuming that the temperature and the pressure remain constant, calculate the volume of air at 1 atm, 20oC, necessary to dilute the HNO3(g) to 0.050 ppm(by volume) when 560.g of NO2(g) is allowed to react.

Homework Equations

V=nRT/P where n=number of moles, R=ideal gas constant, T=temperature, P= pressure

The Attempt at a Solution

First calculate the volume of HNO3 gas that will be produced. then using the equation a/b=c/d
I calculated the volume required to dilute

.050L/1,000,000L = 292.808L / V

solve for V and I got 5.9 x 10^9 L

 

[Borek]

Looks OK.

 

[Blueshift5]

of air

I would like to clarify that the volume of air required to dilute the HNO3 gas to 0.050 ppm is not 5.9 x 10^9 L. This is a very large volume and is not a practical or realistic answer.

To calculate the volume of air required, we need to first calculate the number of moles of HNO3 produced. From the given reaction, we can see that for every 4 moles of NO2, 4 moles of HNO3 are produced. Therefore, the number of moles of HNO3 produced is equal to the number of moles of NO2 given (560 g) divided by the molar mass of NO2 (46.0055 g/mol), which is equal to 12.170 moles.

Next, we can use the ideal gas equation, V=nRT/P, to calculate the volume of air required. We know that the temperature is 20°C, which is equal to 293.15 K, and the pressure is 1 atm. The ideal gas constant, R, is 0.08206 L∙atm/mol∙K. Plugging in these values, we get:

V = (12.170 mol)(0.08206 L∙atm/mol∙K)(293.15 K)/1 atm

V = 297.9 L

Therefore, the volume of air required to dilute the HNO3 gas to 0.050 ppm is 297.9 L. This is a much more reasonable and realistic answer. It is important to be careful with unit conversions and to always check your calculations to ensure that the answer makes sense in the given context.