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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].
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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