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:
[tex]\dfrac{1\;mole\;of\;NH_4NO_3}{2\;moles\;of\;H_{2}O}[/tex]
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:
[tex]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[/tex]
Now we can use this to determine how many moles of H2O this would produce by using the ration we solved for earlier.
[tex]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[/tex]
And we convert that by getting the molecular mass of H2O, which is 18.02 g/mole:
[tex]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[/tex]
But this is only if the whole 45.7 g of ammonium nitrate is used up.
