Answer:
[tex]\large \boxed{\text{60.0 g NH$_{4}$NO$_{3}$; 27.0 g of H$_{2}$O}}[/tex]
Explanation:
We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
MM: 80.04 44.01 18.02
NH₄NO₃ ⟶ N₂O + 2H₂O
m/g: 33.0
1. Mass of NH₄NO₃
(a) Moles of N₂O
[tex]\text{Moles of N$_{2}$O} = \text{33.0 g N$_{2}$O}\times \dfrac{\text{1 mol N$_{2}$O}}{\text{44.01 g N$_{2}$O}}= \text{0.7498 mol N$_{2}$O}[/tex]
(b) Moles of NH₄NO₃
[tex]\text{Moles of NH$_{4}$NO$_{3}$} = \text{0.7498 mol N$_{2}$O} \times \dfrac{\text{1 mol NH$_{4}$NO$_{3}$}}{\text{1 mol N$_{2}$O}} = \text{0.7498 mol NH$_{4}$NO$_{3}$}[/tex]
(c) Mass of NH₄NO₃
[tex]\text{Mass of NH$_{4}$NO$_{3}$} =\text{0.7498 mol NH$_{4}$NO$_{3}$} \times \dfrac{\text{80.04 g NH$_{4}$NO$_{3}$}}{\text{1 mol NH$_{4}$NO$_{3}$}} = \textbf{60.0 g NH$_{4}$NO$_{3}$}[/tex]
2. Mass of H₂O
(a) Moles of H₂O
[tex]\text{Moles of H$_{2}$O} = \text{0.7498 mol N$_{2}$O} \times \dfrac{\text{2 mol H$_{2}$O}}{\text{1 mol N$_{2}$O}} = \text{1.500 mol H$_{2}$O}[/tex]
(c) Mass of H₂O
[tex]\text{Mass of H$_{2}$O} =\text{1.500 mol H$_{2}$O} \times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \textbf{27.0 g H$_{2}$O}\\\\\text{The reaction requires = $\large \boxed{\textbf{60.0 g NH$_{4}$NO$_{3}$}}$ and produces $\large \boxed{\textbf{27.0 g of H$_{2}$O}}$}[/tex]
