Question

Ammonium nitrate decomposes to dinitrogen monoxide and water. If given 45.7 grams of ammonium

Answer 1

Answer:

20.54 g of H₂O

Explanation:

Since you already have a balanced equation, the next step is to see the ratio between the ammonium nitrate and water in the equation:

NH4NO3(s)—N2O(g)+2H2O

1 mole of ammonium nitrate produces 2 moles of H2O

So we have the ration:

$\dfrac{1\;mole\;of\;NH_4NO_3}{2\;moles\;of\;H_{2}O}$

Let's leave that for later use.

Next step is to covert the mass given into moles. We do that by getting the molar mass of the given and using that as a conversion factor:

Element   number of                 molar mass

                   atoms                  of each element          

N =                  2              x            14.01 g/mole   =      28.02 g/mole

H =                  4              x              1.01 g/mole   =        4.01  g/mole

O =                  3              x            16.00 g/mole =      48.00 g/mole

                                                                                    80 .03 g/mole

Now we can convert:

$45.7g\;of\;NH_4NO_3\times\dfrac{1\;of\;NH_4NO_3}{80.03\;g\;of\;NH_4NO_3}=0.57\;moles\;of\;NH_4NO_3$

Now we can use this to determine how many moles of H2O this would produce by using the ration we solved for earlier.

$0.57\;moles\;of\;NH_4NO_3\times\dfrac{2\;moles\;of\;H_{2}O}{1\;mole\;of\;NH_4NO_3} =1.14\;moles\;of\;H_{2}O$

And we convert that by getting the molecular mass of H2O, which is 18.02 g/mole:

$1.14\;moles\;of\;H_{2}O\times\dfrac{18.02\;g\;of\;H_{2}O}{1\;\;mole\;of\;H_{2}O} =20.54\;g\;of\;H_{2}O$

But this is only if the whole 45.7 g of ammonium nitrate is used up.