Answer:
[tex]\large \boxed{\text{11.9 g}}[/tex]
Explanation:
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 17.03 18.02
4NH₃ + 6NO ⟶ 5N₂ + 3H₂O
m/g: 15.0
(a) Moles of NH₃
[tex]\text{Moles of NH}_{3} = \text{15.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{0.8808 mol NH}_{3}[/tex]
(b) Moles of H₂O
[tex]\text{Moles of H$_{2}$O} = \text{0.8808 mol NH}_{3} \times \dfrac{\text{3 mol H$_{2}$O}}{\text{4 mol NH}_{3}} = \text{0.6606 mol H$_{2}$O}[/tex]
(c) Mass of H₂O
[tex]\text{Mass of H$_{2}$O} =\text{0.6606 mol H$_{2}$O} \times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \textbf{11.9 g H$_{2}$O}\\\\\text{The reaction will produce $\large \boxed{\textbf{11.9 g}}$ of H$_{2}$O.}[/tex]
