Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
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.
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.