To determine the mass of water ([tex]\(H_2O\)[/tex]) produced from the decomposition of ammonium nitrate ([tex]\(NH_4NO_3\)[/tex]), we will follow a step-by-step stoichiometric process:
1. Calculate the number of moles of [tex]\(NH_4NO_3\)[/tex]:
Given:
- Mass of [tex]\(NH_4NO_3\)[/tex] = 160.1 grams
- Molar mass of [tex]\(NH_4NO_3\)[/tex] = 80.03 g/mol
Number of moles of [tex]\(NH_4NO_3\)[/tex] is calculated as:
[tex]\[
\text{moles of } NH_4NO_3 = \frac{\text{mass of } NH_4NO_3}{\text{molar mass of } NH_4NO_3}
\][/tex]
[tex]\[
\text{moles of } NH_4NO_3 = \frac{160.1 \text{ grams}}{80.03 \text{ g/mol}} \approx 2.0005 \text{ moles}
\][/tex]
2. Use the stoichiometry of the reaction to find the number of moles of [tex]\(H_2O\)[/tex] produced:
The balanced chemical equation is:
[tex]\[
NH_4NO_3 \rightarrow N_2O + 2 H_2O
\][/tex]
According to the stoichiometry, 1 mole of [tex]\(NH_4NO_3\)[/tex] produces 2 moles of [tex]\(H_2O\)[/tex]. Therefore:
[tex]\[
\text{moles of } H_2O = \text{moles of } NH_4NO_3 \times 2
\][/tex]
[tex]\[
\text{moles of } H_2O = 2.0005 \text{ moles} \times 2 \approx 4.001 \text{ moles}
\][/tex]
3. Calculate the mass of [tex]\(H_2O\)[/tex] produced:
Given:
- Molar mass of [tex]\(H_2O\)[/tex] = 18.01 g/mol
Mass of [tex]\(H_2O\)[/tex] is calculated as:
[tex]\[
\text{mass of } H_2O = \text{moles of } H_2O \times \text{molar mass of } H_2O
\][/tex]
[tex]\[
\text{mass of } H_2O = 4.001 \text{ moles} \times 18.01 \text{ g/mol} \approx 72.06 \text{ grams}
\][/tex]
Therefore, the mass of [tex]\(H_2O\)[/tex] produced is:
[tex]\(\boxed{72.06 \, \text{grams}}\)[/tex]
