From the brand-new conditions of #STP#, the brand-new molar volume transforms out to be #22.7 \ "L"#.

You are watching: Ammonium nitrate decomposes to nitrogen gas and water

Here, us must produce #30 \ "L"# of oxygen gas, and also so we must an initial convert the amount right into moles.

#(30color(red)cancelcolor(black)"L")/(22.7color(red)cancelcolor(black)"L""/mol")~~1.32 \ "mol"#

In other words, we require to produce #1.32# moles of oxygen.

The mole ratio between #NH_4NO_3# and also #O_2# is #2:1#. So, #2# mole of ammonium nitrate are essential to develop one mole that oxygen.

So, us would require #1.32xx2=2.64# mole of ammonium nitrate to develop #1.32# mole of oxygen.

Ammonium nitrate #(NH_4NO_3)# has a molar mass of #80.043 \ "g/mol"#.

So here, we will certainly need

#2.64color(red)cancelcolor(black)"mol"*(80.043 \ "g")/(1color(red)cancelcolor(black)"mol")=211.31 \ "g"#

So, about #211.31# grams the ammonium nitrate would certainly be essential to create #30# liters that oxygen gas in ~ #STP#.

Truong-Son N.
Mar 5, 2018

Why would we assume STP? just state the temperature. A rapid google search provides a usual temperature for this reaction the #300^
"C"#, and a minimum temperature of #210^
"C"#. In ~ that preferred temperature that #300^
"C"#, I obtain #"100.8 g"# is needed.

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If for some reason I would pick the unphysical temperature that #0^
"C"#, then I would certainly hypothetically create #"1.321 mols O"_2# indigenous #"2.642 mols NH"_4"NO"_3# (it wouldn"t in reality happen...), which would correspond come #"211.4 g"#...

#2"NH"_4"NO"_3(s) -> 2"N"_2(g) + 4"H"_2"O"(g) + "O"_2(g)#

If you desire to do #"30 L"# #"O"_2# gas (I will still i think a standard pressure of #"1 bar"#), climate that corresponds to

#V_(O_2) = (n_(O_2)RT)/P^
#

So, there are this countless mols the form:

#n_(O_2) = (P^
V_(O_2))/(RT)#

#= ("1 bar" cdot "30 L")/("0.083145 L"cdot"bar/mol"cdot"K" cdot "573.15 K")#

#= "0.6295 mols O"_2#

Since #"2 mols"# that #"NH"_4"NO"_3# to produce #"O"_2# (assuming 100% yield), us would need at least #2 xx 0.6295 = "1.259 mols NH"_4"NO"_3# to begin with to type #"0.6295 mols O"_2#.

Therefore, we need this plenty of grams:

#1.259 cancel("mols NH"_4"NO"_3) xx ("80.0426 g")/cancel("1 mol NH"_4"NO"_3)#

#= color(blue)("100.8 g NH"_4"NO"_3)#