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Ammonium perchlorate [tex] \left( \text{NH}_4 \text{ClO}_4 \right) [/tex] is the solid rocket fuel that was used by the U.S. Space Shuttle and is used in the Space Launch System (SLS) of the Artemis rocket. It reacts with itself to produce nitrogen gas [tex] \left( \text{N}_2 \right) [/tex], chlorine gas [tex] \left( \text{Cl}_2 \right) [/tex], oxygen gas [tex] \left( \text{O}_2 \right) [/tex], water [tex] \left( \text{H}_2 \text{O} \right) [/tex], and a great deal of energy.

What mass of oxygen gas is produced by the reaction of 3.62 g of ammonium perchlorate?

Round your answer to 3 significant digits.

[tex]$\square$[/tex]

Sagot :

GinnyAnswer
To determine the mass of oxygen gas produced from the reaction of 3.62 g of ammonium perchlorate ([tex]\( \text{NH}_4\text{ClO}_4 \)[/tex]), follow these steps:

1. Calculate the molar mass of ammonium perchlorate ([tex]\( \text{NH}_4\text{ClO}_4 \)[/tex]):
- For Nitrogen (N): 1 atom × 14.01 g/mol = 14.01 g/mol
- For Hydrogen (H): 4 atoms × 1.01 g/mol = 4.04 g/mol
- For Chlorine (Cl): 1 atom × 35.45 g/mol = 35.45 g/mol
- For Oxygen (O): 4 atoms × 16.00 g/mol = 64.00 g/mol

Sum these together:
[tex]\[ \text{Molar mass of} \ \text{NH}_4\text{ClO}_4 = 14.01 + 4.04 + 35.45 + 64.00 = 117.49 \ \text{g/mol} \][/tex]

2. Calculate the moles of ammonium perchlorate:
[tex]\[ \text{Moles of} \ \text{NH}_4\text{ClO}_4 = \frac{\text{Mass of} \ \text{NH}_4\text{ClO}_4}{\text{Molar mass of} \ \text{NH}_4\text{ClO}_4} = \frac{3.62 \ \text{g}}{117.49 \ \text{g/mol}} = 0.030811132862371266 \ \text{mol} \][/tex]

3. Determine the stoichiometry of the reaction:
The balanced chemical equation for the decomposition of ammonium perchlorate is:
[tex]\[ 2 \text{NH}_4\text{ClO}_4 \rightarrow N_2 + 2 \text{Cl}_2 + 2 \text{O}_2 + 4 \text{H}_2\text{O} \][/tex]

From this reaction, we see that 2 moles of [tex]\( \text{NH}_4\text{ClO}_4 \)[/tex] produce 2 moles of [tex]\( \text{O}_2 \)[/tex]. This means the ratio is 1:1 for [tex]\( \text{NH}_4\text{ClO}_4 \)[/tex] to [tex]\( \text{O}_2 \)[/tex].

4. Calculate the moles of oxygen gas ([tex]\( \text{O}_2 \)[/tex]) produced:
Since the stoichiometric ratio is 1:1,
[tex]\[ \text{Moles of} \ \text{O}_2 \ \text{produced} = 0.030811132862371266 \ \text{mol} \][/tex]

5. Calculate the mass of the oxygen gas produced:
The molar mass of [tex]\( \text{O}_2 \)[/tex] (oxygen gas) is 32.00 g/mol (since 2 atoms of oxygen, each with a molar mass of 16.00 g/mol).

[tex]\[ \text{Mass of} \ \text{O}_2 = \text{Moles of} \ \text{O}_2 \times \text{Molar mass of} \ \text{O}_2 \][/tex]

[tex]\[ \text{Mass of} \ \text{O}_2 = 0.030811132862371266 \ \text{mol} \times 32.00 \ \text{g/mol} = 0.9859562515958805 \ \text{g} \][/tex]

6. Round the result to 3 significant digits:
[tex]\[ \text{Mass of} \ \text{O}_2 \ \text{produced} = 0.986 \ \text{g} \][/tex]

Therefore, the mass of oxygen gas produced by the reaction of 3.62 g of ammonium perchlorate is [tex]\( \boxed{0.986} \ \text{grams} \)[/tex].
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