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The chemical equation below shows the decomposition of ammonium nitrate [tex]$\left( NH_4NO_3 \right)$[/tex]:
[tex]$
NH_4NO_3 \rightarrow N_2O + 2H_2O
$[/tex]

A chemist who is performing this reaction starts with 160.1 g of [tex]$NH_4NO_3$[/tex]. The molar mass of [tex]$NH_4NO_3$[/tex] is [tex]$80.03 \, \text{g/mol}$[/tex]; the molar mass of water [tex]$\left( H_2O \right)$[/tex] is [tex]$18.01 \, \text{g/mol}$[/tex].

What mass, in grams, of [tex]$H_2O$[/tex] is produced?

A. 9.01 grams
B. 18.01 grams
C. 36.03 grams
D. 72.06 grams

Sagot :

GinnyAnswer
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]
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