320 g of solid ammonium nitrite, NH_4NO_2, decomposes when heated according to the balanced equation: NH_4NO_2 -> N_2+2H_2O. What total volume of gases at 819 K is emitted by this reaction?

1 Answer
David G.
Mar 21, 2016

Each mole of NH_4NO_2 produces 3 mol of gases total (N_2 and H_2O), and we have 5 mol. So 3xx5 = 15 mol of gas at STP is 336 L. Converting to 819 K gives a volume of 1008 L.

Explanation:

The molar mass of NH_4NO_2 is 64 gmol^-1 (two N at 14, four H at 1, two O at 16).

Find the number of moles of NH_4NO_2:

n=m/M=320/64=5 mol

From the balanced equation, each mole of NH_4NO_2 yields 1 mole of N_2 and 2 moles of H_2O (which is definitely a gas at 819 K), for a total of 3 moles of gas.

For ideal gases (which we can treat these real gases as for our purposes), 1 mole of any gas at STP (273 K and 1 atm) occupies 22.4 L.

That means we have 3xx5 = 15 mol of product gases, and they occupy a volume of 336 L at STP.

We need to change the temperature from 273 K to 819 K, and the pressure stays the same at 1 atm, so:

V_1/T_1=V_2/T_2

Rearranging:

V_2=V_1T_2/T_1 = 336 xx 819/273 = 1008 L

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