Ammonium nitrate decomposes to dinitrogen monoxide and water. If given 45.7 grams of ammonium
nitrate, how many grams of water would be produced?
NH4NO3(s)—N2O(g)+2H20(1)
0

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:

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:

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

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

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

the answer is 1 i believe