Ammonium nitrate will decompose explosively at high temperatures to form nitrogen, oxygen, and water vapor. 2NH4NO3(s) - - > 2N2(g) + 4H2O(g) + O2(g) How many grams of NH4NO 3 are needed to produce 30 L of O 2 gas? (Assume STP.)

We have the balanced equation:

#2NH_4NO_3(s)stackrel(Delta)->2N_2(g)+4H_2O(g)+O_2(g)#

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

Here, we must produce #30 \ "L"# of oxygen gas, and so we must first convert that amount into moles.

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

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

The mole ratio between #NH_4NO_3# and #O_2# is #2:1#. So, #2# moles of ammonium nitrate are needed to produce one mole of oxygen.

So, we would need #1.32xx2=2.64# moles of ammonium nitrate to produce #1.32# moles of oxygen.

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

So here, we will need

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

So, approximately #211.31# grams of ammonium nitrate would be needed to produce #30# liters of oxygen gas at #STP#.

Why would we assume STP? Just state the temperature. A quick google search gives a typical temperature for this reaction of #300^@ "C"#, and a minimum temperature of #210^@ "C"#. At that chosen temperature of #300^@ "C"#, I get #"100.8 g"# is needed.

If for some reason I would pick the unphysical temperature of #0^@ "C"#, then I would hypothetically produce #"1.321 mols O"_2# from #"2.642 mols NH"_4"NO"_3# (it wouldn't actually happen...), which would correspond to #"211.4 g"#...

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

If you want to make #"30 L"# #"O"_2# gas (I will still assume a standard pressure of #"1 bar"#), then that corresponds to

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

So, there are this many mols that 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"# of #"NH"_4"NO"_3# produces #"O"_2# (assuming 100% yield), we would need at least #2 xx 0.6295 = "1.259 mols NH"_4"NO"_3# to begin with to form #"0.6295 mols O"_2#.

Therefore, we need this many 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)#