Hydrogen cyanide gas (HCN, 27.03 g/mol) is extremely toxic.Levels around 3.2 mg/L can kill a person in about one minute.Suppose a reaction generates 0.0345 atm of HCN gas at 0.00 $\degree$C in a 15.0 L fume hood.How many milligrams per liter of HCN is this?

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To find the concentration of HCN in mg/L, we can use the ideal gas law equation PV = nRT, where P is the pressure in atm (0.0345 atm), V is the volume in liters (15.0 L), n is the number of moles of gas, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin. First, convert the temperature from Celsius to Kelvin (0°C + 273.15 = 273.15 K). Then, solve for n (moles of HCN) using the ideal gas law equation. Finally, convert moles of HCN to grams (using the molar mass of HCN, 27.03 g/mol) and then to milligrams, and divide by the volume in liters to get mg/L.n = (P*V) / (R*T) = (0.0345 atm * 15.0 L) / (0.0821 L·atm/(mol·K) * 273.15 K) = 0.0231 molMass of HCN = n * molar mass = 0.0231 mol * 27.03 g/mol = 0.624 gConvert grams to milligrams = 0.624 g * 1000 mg/g = 624 mgSince the volume of the fume hood is 15.0 L, the concentration of HCN in mg/L is 624 mg / 15.0 L = 41.6 mg/L.

Hydrogen Cyanide Gas (HCN, 27 °\degree° C in a 150 L Fume Hood

Hydrogen Cyanide Gas (HCN, 27 $\degree$ C in a 150 L Fume Hood