Answer :
To determine how many moles of [tex]\( NO_2 \)[/tex] form when [tex]\( 63.25 \, \text{g} \, N_2O_5 \)[/tex] decompose, follow these detailed steps:
1. Determine the molar mass of [tex]\( N_2O_5 \)[/tex]:
- The molar mass of Nitrogen ([tex]\( N \)[/tex]) is [tex]\( 14 \, \text{g/mol} \)[/tex].
- The molar mass of Oxygen ([tex]\( O \)[/tex]) is [tex]\( 16 \, \text{g/mol} \)[/tex].
- [tex]\( N_2O_5 \)[/tex] consists of 2 Nitrogen atoms and 5 Oxygen atoms.
- Calculate the molar mass of [tex]\( N_2O_5 \)[/tex]:
[tex]\[ \text{Molar mass of } N_2O_5 = 2(14) + 5(16) = 28 + 80 = 108 \, \text{g/mol} \][/tex]
2. Calculate the number of moles of [tex]\( N_2O_5 \)[/tex]:
- Use the formula:
[tex]\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} \][/tex]
- Given the mass of [tex]\( N_2O_5 \)[/tex] is [tex]\( 63.25 \, \text{g} \)[/tex]:
[tex]\[ \text{Number of moles of } N_2O_5 = \frac{63.25 \, \text{g}}{108 \, \text{g/mol}} = 0.5856481481481481 \, \text{mol} \][/tex]
3. Use the stoichiometry of the reaction:
- According to the balanced chemical equation:
[tex]\[ 2 N_2O_5 \rightarrow 4 NO_2 + O_2 \][/tex]
- 2 moles of [tex]\( N_2O_5 \)[/tex] produce 4 moles of [tex]\( NO_2 \)[/tex].
- Therefore, 1 mole of [tex]\( N_2O_5 \)[/tex] will produce 2 moles of [tex]\( NO_2 \)[/tex].
- Using the number of moles of [tex]\( N_2O_5 \)[/tex] calculated:
4. Calculate the moles of [tex]\( NO_2 \)[/tex] produced:
[tex]\[ \text{Number of moles of } NO_2 = 0.5856481481481481 \, \text{mol of } N_2O_5 \times 2 = 1.1712962962962963 \, \text{mol of } NO_2 \][/tex]
Therefore, when [tex]\( 63.25 \, \text{g} \, N_2O_5 \)[/tex] decomposes, [tex]\( 1.1712962962962963 \, \text{mol} \, NO_2 \)[/tex] are formed.
1. Determine the molar mass of [tex]\( N_2O_5 \)[/tex]:
- The molar mass of Nitrogen ([tex]\( N \)[/tex]) is [tex]\( 14 \, \text{g/mol} \)[/tex].
- The molar mass of Oxygen ([tex]\( O \)[/tex]) is [tex]\( 16 \, \text{g/mol} \)[/tex].
- [tex]\( N_2O_5 \)[/tex] consists of 2 Nitrogen atoms and 5 Oxygen atoms.
- Calculate the molar mass of [tex]\( N_2O_5 \)[/tex]:
[tex]\[ \text{Molar mass of } N_2O_5 = 2(14) + 5(16) = 28 + 80 = 108 \, \text{g/mol} \][/tex]
2. Calculate the number of moles of [tex]\( N_2O_5 \)[/tex]:
- Use the formula:
[tex]\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} \][/tex]
- Given the mass of [tex]\( N_2O_5 \)[/tex] is [tex]\( 63.25 \, \text{g} \)[/tex]:
[tex]\[ \text{Number of moles of } N_2O_5 = \frac{63.25 \, \text{g}}{108 \, \text{g/mol}} = 0.5856481481481481 \, \text{mol} \][/tex]
3. Use the stoichiometry of the reaction:
- According to the balanced chemical equation:
[tex]\[ 2 N_2O_5 \rightarrow 4 NO_2 + O_2 \][/tex]
- 2 moles of [tex]\( N_2O_5 \)[/tex] produce 4 moles of [tex]\( NO_2 \)[/tex].
- Therefore, 1 mole of [tex]\( N_2O_5 \)[/tex] will produce 2 moles of [tex]\( NO_2 \)[/tex].
- Using the number of moles of [tex]\( N_2O_5 \)[/tex] calculated:
4. Calculate the moles of [tex]\( NO_2 \)[/tex] produced:
[tex]\[ \text{Number of moles of } NO_2 = 0.5856481481481481 \, \text{mol of } N_2O_5 \times 2 = 1.1712962962962963 \, \text{mol of } NO_2 \][/tex]
Therefore, when [tex]\( 63.25 \, \text{g} \, N_2O_5 \)[/tex] decomposes, [tex]\( 1.1712962962962963 \, \text{mol} \, NO_2 \)[/tex] are formed.
