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

Using the following * unbalanced * equation, answer the following questions:

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

a.) What is the mole ratio of NH4NO3 to H2O?

b.) How many grams of water are produced if 33.0 grams of N2O is produced?

c.) How many moles of N2O are produced if you have 25 grams of NH4NO3?

d.) How many moles of H2O are produced if you have 5 moles of NH4NO3?

e.) How many grams of N2O are produced if you have 125 grams of NH4NO3?

1 Answer
Truong-Son N.
Mar 19, 2018

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")#

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