Just how dangerous is NH3?

In my Process Safety, HAZMAT, and Emergency Response course, I always walk the students through an exercise to demonstrate just how dangerous chemicals such as Cl2 and NH3 are. So many have come to believe that these chemicals are hazardous only when we exceed the OSHA/EPA Thresholds for PSM/RMP.

In a 40 ft × 40 ft space with an 8ft ceiling, we need only 111.5 mL (0.0303798 gallons) of liquid NH3 to make the room an IDLH Atmosphere (e.g., 300 ppm). A far cry from OSHA’s and EPA’s PSM/RMP TQ of 10,000 pounds.

How much liquid NH3 does it take to achieve 300 ppm in a room that is 40′ X 40′ X8′?

So I’ll explain the theoretical chemistry, but not how to generate chlorine gas in a real space.

How much Liquid NH3 is needed to make a room 40’X40’X8′ reach 300 ppm?

Short answer:

Room volume in cubic meters:

    Dimensions: 40′ × 40′ × 8′

    Volume in cubic feet: 40 × 40 × 8 = 12,800 ft3

    Convert to cubic meters using 1 ft3≈0.0283168 m3:

    12,800 × 0.0283168 ≈ 362.45 m3

    What 300 ppm means:

      300 ppm by volume means

      300/1,000,000 = 0.0003 of the room’s volume is ammonia gas.

      So ammonia gas volume:
      0.0003 × 362.45 ≈ 0.1087m3

      Convert that gas volume to moles and mass:

        Assume near room conditions, molar volume ≈24.45 L/mol = 0.02445 m3/mol.

        Moles of NH3:

        n = 0.10870.02445 ≈ 4.45 mol

        Molar mass of NH3 ≈ 17.03 g/mol.

        m = 4.45×17.03 ≈ 75.8 g

        That’s the mass if the whole room were filled with that 0.1087m³ of pure ammonia. But we already accounted for ppm via volume fraction, so this 75.8g is the mass of ammonia corresponding to 300 ppm in the whole 362.45m³.

        However, using the standard ppm↔mg/m³ formula for gases is cleaner:

        For ammonia, at ~25°C/77°F:

        mg/m3 = ppm × molar mass 24.45 =300 × 17.0324.45 ≈ 209 mg/m3

        Total mass in the room:

        209 × 362.45 ≈ 75,800mg ≈ 75.8 g

        Convert gas mass to liquid volume:

          The density of liquid ammonia at around its boiling point is about 0.68 g/mL.

          V liquid=75.8 g0.68 g/mL ≈ 111.5 mL

          So the consistent result is: you’d need about 76 grams of ammonia, which is about 110 mL of liquid NH3, to reach 300 ppm in that room, assuming:
          1) It fully vaporizes,
          2) Mixes uniformly,
          3) Conditions are near room temperature and 1 atm.

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