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To determine how many moles of [tex]\(N_2O_5\)[/tex] are needed to produce 7.90 grams of [tex]\(NO_2\)[/tex], we need to take a systematic approach. Follow these steps:
1. Calculate the Molar Mass of [tex]\(NO_2\)[/tex]:
- The molar mass of [tex]\(NO_2\)[/tex] can be found by adding the atomic masses of nitrogen (N) and oxygen (O).
- Nitrogen (N) has an atomic mass of approximately 14.007 g/mol, and oxygen (O) has an atomic mass of approximately 15.999 g/mol.
- [tex]\( NO_2 \)[/tex] consists of 1 nitrogen and 2 oxygens:
[tex]\[ \text{Molar mass of } NO_2 = 14.007 + (2 \times 15.999) = 46.0055 \text{ g/mol} \][/tex]
2. Calculate the Number of Moles of [tex]\(NO_2\)[/tex]:
- Using the molar mass of [tex]\(NO_2\)[/tex], we can convert the given mass of [tex]\(NO_2\)[/tex] to moles.
- Given mass of [tex]\(NO_2\)[/tex] is 7.90 grams.
- The number of moles of [tex]\(NO_2\)[/tex] can be calculated using the formula:
[tex]\[ \text{Moles of } NO_2 = \frac{\text{Mass of } NO_2}{\text{Molar mass of } NO_2} \][/tex]
[tex]\[ \text{Moles of } NO_2 = \frac{7.90 \text{ g}}{46.0055 \text{ g/mol}} \approx 0.1717 \text{ moles} \][/tex]
3. Utilize Stoichiometry from the Balanced Equation:
- According to the balanced chemical equation:
[tex]\[ 2 N_2O_5 \rightarrow 4 NO_2 + O_2 \][/tex]
- This indicates that 2 moles of [tex]\(N_2O_5\)[/tex] produce 4 moles of [tex]\(NO_2\)[/tex].
- Therefore, the mole ratio of [tex]\(N_2O_5\)[/tex] to [tex]\(NO_2\)[/tex] is [tex]\(\frac{2}{4}\)[/tex] or [tex]\(\frac{1}{2}\)[/tex].
4. Calculate the Moles of [tex]\(N_2O_5\)[/tex] Needed:
- Using the mole ratio, we can find the number of moles of [tex]\(N_2O_5\)[/tex] needed to produce the given moles of [tex]\(NO_2\)[/tex].
- The relationship is:
[tex]\[ \text{Moles of } N_2O_5 = \text{Moles of } NO_2 \times \left(\frac{\text{Moles of } N_2O_5}{\text{Moles of } NO_2}\right) \][/tex]
- From this:
[tex]\[ \text{Moles of } N_2O_5 = 0.1717 \text{ moles} \times \left(\frac{1}{2}\right) \approx 0.0859 \text{ moles} \][/tex]
In conclusion, to produce 7.90 grams of [tex]\(NO_2\)[/tex], approximately 0.0859 moles of [tex]\(N_2O_5\)[/tex] are needed.
1. Calculate the Molar Mass of [tex]\(NO_2\)[/tex]:
- The molar mass of [tex]\(NO_2\)[/tex] can be found by adding the atomic masses of nitrogen (N) and oxygen (O).
- Nitrogen (N) has an atomic mass of approximately 14.007 g/mol, and oxygen (O) has an atomic mass of approximately 15.999 g/mol.
- [tex]\( NO_2 \)[/tex] consists of 1 nitrogen and 2 oxygens:
[tex]\[ \text{Molar mass of } NO_2 = 14.007 + (2 \times 15.999) = 46.0055 \text{ g/mol} \][/tex]
2. Calculate the Number of Moles of [tex]\(NO_2\)[/tex]:
- Using the molar mass of [tex]\(NO_2\)[/tex], we can convert the given mass of [tex]\(NO_2\)[/tex] to moles.
- Given mass of [tex]\(NO_2\)[/tex] is 7.90 grams.
- The number of moles of [tex]\(NO_2\)[/tex] can be calculated using the formula:
[tex]\[ \text{Moles of } NO_2 = \frac{\text{Mass of } NO_2}{\text{Molar mass of } NO_2} \][/tex]
[tex]\[ \text{Moles of } NO_2 = \frac{7.90 \text{ g}}{46.0055 \text{ g/mol}} \approx 0.1717 \text{ moles} \][/tex]
3. Utilize Stoichiometry from the Balanced Equation:
- According to the balanced chemical equation:
[tex]\[ 2 N_2O_5 \rightarrow 4 NO_2 + O_2 \][/tex]
- This indicates that 2 moles of [tex]\(N_2O_5\)[/tex] produce 4 moles of [tex]\(NO_2\)[/tex].
- Therefore, the mole ratio of [tex]\(N_2O_5\)[/tex] to [tex]\(NO_2\)[/tex] is [tex]\(\frac{2}{4}\)[/tex] or [tex]\(\frac{1}{2}\)[/tex].
4. Calculate the Moles of [tex]\(N_2O_5\)[/tex] Needed:
- Using the mole ratio, we can find the number of moles of [tex]\(N_2O_5\)[/tex] needed to produce the given moles of [tex]\(NO_2\)[/tex].
- The relationship is:
[tex]\[ \text{Moles of } N_2O_5 = \text{Moles of } NO_2 \times \left(\frac{\text{Moles of } N_2O_5}{\text{Moles of } NO_2}\right) \][/tex]
- From this:
[tex]\[ \text{Moles of } N_2O_5 = 0.1717 \text{ moles} \times \left(\frac{1}{2}\right) \approx 0.0859 \text{ moles} \][/tex]
In conclusion, to produce 7.90 grams of [tex]\(NO_2\)[/tex], approximately 0.0859 moles of [tex]\(N_2O_5\)[/tex] are needed.
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