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11.5 (1)

11.5 (1) - equation 2NaN 3 → 2Na 3N 2(don’t forget that...

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Key: (Try this without a calculator) What volume of nitrogen gas is produced if 8 g of NaN 3 decomposes at 127 °C to form sodium metal and nitrogen gas at 1 atm pressure? By the way, this is the reaction in an auto air bag. a. 1 L b. 0.1 L c. 2.5 L d. 4 L e. 9 L f. 6 L This question asks what volume of nitrogen gas is produced in the reaction when you start with 8 grams of reactant. Remember, any time you are asked to convert between units like grams, moles or liters, you need to use dimensional analysis. Think about the conversions that you will need to make. Here is a scheme that will help: g NaN 3 → moles NaN 3 →moles N 2 →liters N 2 . For the first conversion you need to know the molar mass of NaN 3. You can determine this from the periodic table. MM NaN 3 = 65g. To convert from moles NaN 3 to moles N 2 you need to know their stoichiometric ratios by looking at the balanced equation. Since it is not provided, you need to come up with it. Balanced
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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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