#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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