Unformatted text preview: equation: 2NaN 3 → 2Na + 3N 2 . (don’t forget that nitrogen gas is always diatomic!). For the last conversion, moles N 2 →liters N 2 , you need to know the volume of one mole of N 2 gas. At STP, this number is 22.4 for all gases. Since this problem is not at STP, you need to use the ideal gas law (PV = nRT) to determine the volume of one mole of gas. R is a constant that equals 0.082 L-atm/Kmol. P and T are provided in the problem, and we want to know the volume of gas in one mole, so we can make n = 1. Substituting in the provided information we have 1atmV = 1mol(0.082)(127 +273K) which equals about 32L. (Remember, you always want to use temperature in K units.) We can now use dimensional analysis to find the volume of N 2 using the mass of NaN 3 provided: 8g NaN 3 │1mol NaN 3 │3mol N 2 │32L N 2 = 6 L 65g NaN 3 │2mol NaN 3 │1mol N 2...
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- Fall '08
- Reaction, 4 L, 1 L, 6 L, nitrogen gas