Calculating Volume of Air for Nitrogen Dioxide Dillution

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SUMMARY

The discussion focuses on calculating the volume of air required to dilute nitric acid (HNO3) gas produced from the reaction of nitrogen dioxide (NO2) with water. The reaction is represented by the equation 4NO2(g) + 2H2O(g) + O2(g) → 4HNO3(g). Using the ideal gas law equation V = nRT/P, the user calculated the volume of HNO3 gas produced and determined that to achieve a dilution of 0.050 ppm, approximately 5.9 x 10^9 liters of air at 1 atm and 20°C is necessary.

PREREQUISITES
  • Understanding of stoichiometry and chemical reactions
  • Familiarity with the ideal gas law (V = nRT/P)
  • Knowledge of parts per million (ppm) calculations
  • Basic principles of gas behavior under constant temperature and pressure
NEXT STEPS
  • Study the ideal gas law in detail, including its applications and limitations
  • Learn about stoichiometric calculations in chemical reactions
  • Research methods for calculating gas volumes in atmospheric chemistry
  • Explore the environmental impact of nitrogen dioxide and acid rain formation
USEFUL FOR

Chemistry students, environmental scientists, and anyone involved in atmospheric studies or pollution control will benefit from this discussion.

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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
 
Last edited:
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Looks OK.
 

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