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To determine how many molecules of nitrogen monoxide (NO) would form if [tex]\(6.3 \times 10^{25}\)[/tex] formula units of aluminum oxide (Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex]) were produced, we will go through the steps of conversion from formula units to molecules using stoichiometry principles. Here is the detailed, step-by-step method:
### Step 1: Convert Formula Units of Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex] to Moles
First, we need to convert the given formula units of Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex] to moles. We use Avogadro's number for this conversion. Avogadro's number ([tex]\(N_A\)[/tex]) is [tex]\(6.022 \times 10^{23}\)[/tex] formula units per mole.
[tex]\[ \text{Moles of Al}_2\text{O}_3 = \frac{\text{Formula units of Al}_2\text{O}_3}{N_A} \][/tex]
Substitute the given values:
[tex]\[ \text{Moles of Al}_2\text{O}_3 = \frac{6.3 \times 10^{25}}{6.022 \times 10^{23}} = 104.6164065094653 \text{ moles} \][/tex]
### Step 2: Use the Mole Ratio from the Balanced Equation
Next, we need to use the mole ratio between Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex] and NO from the balanced chemical equation. The balanced equation gives the relationship between the reactants and products. Assuming the balanced equation for the reaction is:
[tex]\[ 2 \text{Al}_2\text{O}_3 + 6 \text{NO} \rightarrow \text{products} \][/tex]
From this equation, we see that 2 moles of Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex] produce 6 moles of NO.
Thus, the mole ratio of NO to Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex] is:
[tex]\[ \text{Mole ratio} = \frac{6}{2} = 3 \][/tex]
Now, we use this ratio to find the moles of NO produced:
[tex]\[ \text{Moles of NO} = \text{Moles of Al}_2\text{O}_3 \times \text{Mole ratio} \][/tex]
Substitute the values:
[tex]\[ \text{Moles of NO} = 104.6164065094653 \text{ moles} \times 3 = 313.8492195283959 \text{ moles} \][/tex]
### Step 3: Convert Moles of NO to Molecules
Finally, we need to convert the moles of NO to molecules. Again, we use Avogadro's number ([tex]\(6.022 \times 10^{23}\)[/tex]) for this conversion.
[tex]\[ \text{Molecules of NO} = \text{Moles of NO} \times N_A \][/tex]
Substitute the values:
[tex]\[ \text{Molecules of NO} = 313.8492195283959 \text{ moles} \times 6.022 \times 10^{23} = 1.89 \times 10^{26} \text{ molecules} \][/tex]
### Final Answer
When [tex]\(6.3 \times 10^{25}\)[/tex] formula units of aluminum oxide are produced, the number of molecules of nitrogen monoxide (NO) formed is:
[tex]\[ 1.89 \times 10^{26} \text{ molecules} \][/tex]
### Step 1: Convert Formula Units of Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex] to Moles
First, we need to convert the given formula units of Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex] to moles. We use Avogadro's number for this conversion. Avogadro's number ([tex]\(N_A\)[/tex]) is [tex]\(6.022 \times 10^{23}\)[/tex] formula units per mole.
[tex]\[ \text{Moles of Al}_2\text{O}_3 = \frac{\text{Formula units of Al}_2\text{O}_3}{N_A} \][/tex]
Substitute the given values:
[tex]\[ \text{Moles of Al}_2\text{O}_3 = \frac{6.3 \times 10^{25}}{6.022 \times 10^{23}} = 104.6164065094653 \text{ moles} \][/tex]
### Step 2: Use the Mole Ratio from the Balanced Equation
Next, we need to use the mole ratio between Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex] and NO from the balanced chemical equation. The balanced equation gives the relationship between the reactants and products. Assuming the balanced equation for the reaction is:
[tex]\[ 2 \text{Al}_2\text{O}_3 + 6 \text{NO} \rightarrow \text{products} \][/tex]
From this equation, we see that 2 moles of Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex] produce 6 moles of NO.
Thus, the mole ratio of NO to Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex] is:
[tex]\[ \text{Mole ratio} = \frac{6}{2} = 3 \][/tex]
Now, we use this ratio to find the moles of NO produced:
[tex]\[ \text{Moles of NO} = \text{Moles of Al}_2\text{O}_3 \times \text{Mole ratio} \][/tex]
Substitute the values:
[tex]\[ \text{Moles of NO} = 104.6164065094653 \text{ moles} \times 3 = 313.8492195283959 \text{ moles} \][/tex]
### Step 3: Convert Moles of NO to Molecules
Finally, we need to convert the moles of NO to molecules. Again, we use Avogadro's number ([tex]\(6.022 \times 10^{23}\)[/tex]) for this conversion.
[tex]\[ \text{Molecules of NO} = \text{Moles of NO} \times N_A \][/tex]
Substitute the values:
[tex]\[ \text{Molecules of NO} = 313.8492195283959 \text{ moles} \times 6.022 \times 10^{23} = 1.89 \times 10^{26} \text{ molecules} \][/tex]
### Final Answer
When [tex]\(6.3 \times 10^{25}\)[/tex] formula units of aluminum oxide are produced, the number of molecules of nitrogen monoxide (NO) formed is:
[tex]\[ 1.89 \times 10^{26} \text{ molecules} \][/tex]
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