Using the following unbalanced equation, answer the following questions? #"NH"_4"NO"_3(s) stackrel(Delta" ")(->) "N"_2"O"(g) + "H"_2"O"(g)#

Well, as always, we must balance the reaction first...

UNBALANCED:

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

BALANCED:

#"NH"_4"NO"_3(s) stackrel(Delta" ")(->) "N"_2"O"(g) + color(red)2"H"_2"O"(g)#

And now you must review the concepts of extensive and intensive properties. That will greatly decrease the workload...

#a)# The mol ratio can then be readily spotted to be #bb("2 mols H"_2"O"(g))# to #bb("1 mol NH"_4"NO"_3(s))#.

#b)# #"33 g"# of #"N"_2"O"# produced corresponds to

#33 cancel("g N"_2"O") xx ("1 mol N"_2"O")/(2 cdot 14.007 + 15.999 cancel("g N"_2"O")) = ul(??? "mols N"_2"O")#

We see that #"1 mol N"_2"O"# corresponds to #"2 mols H"_2"O"#... so whatever you get there, double it to get the #"mols"# of #"H"_2"O"(g)#...

and so,

#cancel("mols H"_2"O") xx ("18.015 g H"_2"O")/cancel("1 mol H"_2"O") = color(blue)(??? "g H"_2"O"(g))#

#c)# #"25 g"# of #"NH"_4"NO"_3# is this many mols...

#25 cancel("g NH"_4"NO"_3) xx ("1 mol NH"_4"NO"_3)/(2 cdot 14.007 + 4 cdot 1.0079 + 3 cdot 15.999 cancel("g NH"_4"NO"_3))#

#= ul(??? "mols NH"_4"NO"_3) " "stackrel(?" ")(->) " "color(blue)(???"mols N"_2"O")#

If #"1 mol N"_2"O"(g)# was made from #"1 mol"# of #"NH"_4"NO"_3(s)#, do you have to do anything to the #"mols"# you just found to get the #bb"mols"# of #bb("N"_2"O")#?

#d)# Well, do you want to do this the easy way or the hard way?

#5 xx ("NH"_4"NO"_3(s) stackrel(Delta" ")(->) "N"_2"O"(g) + color(red)2"H"_2"O"(g))#

So then, what is #color(blue)(5 xx "2 mols H"_2"O")# produced from #5 xx "1 mol NH"_4"NO"_3# going to give you?

#e)# Well, we knew how many #"mols"# of #"NH"_4"NO"_3# we had in #"25 g"#...

Multiply what you got in part #(c)# by #"125 g"/"25 g" = 5#, and then you will get #5# times as many #"mols"# of #"N"_2"O"# as in part #(c)#. You should be able to explain why we can do this.

#cancel("mols N"_2"O") xx (2 cdot 14.007 + "15.999 g N"_2"O")/cancel("1 mol N"_2"O")#

#= color(blue)(??? "g N"_2"O")#