Join the IDNLearn.com community and start getting the answers you need today. Join our interactive community and access reliable, detailed answers from experienced professionals across a variety of topics.
Sagot :
Sure, let's solve this step-by-step:
1. Determine the moles of aluminum (Al):
- Given mass of aluminum ([tex]\( \text{Al} \)[/tex]) is 18.5 grams.
- The molar mass of aluminum ([tex]\( \text{Al} \)[/tex]) is 26.98 grams per mole.
We find the number of moles of aluminum using the formula:
[tex]\[ \text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} \][/tex]
[tex]\[ \text{moles of Al} = \frac{18.5 \text{ g}}{26.98 \text{ g/mol}} = 0.6856931060044478 \text{ mol} \][/tex]
2. Determine the moles of nitrogen ([tex]\( \text{N}_2 \)[/tex]) required:
- According to the balanced chemical reaction:
[tex]\[ 2 \text{Al} + \text{N}_2 \rightarrow 2 \text{AlN} \][/tex]
2 moles of aluminum ([tex]\( \text{Al} \)[/tex]) react with 1 mole of nitrogen ([tex]\( \text{N}_2 \)[/tex]).
Therefore, the moles of [tex]\( \text{N}_2 \)[/tex] required will be half the moles of [tex]\( \text{Al} \)[/tex]:
[tex]\[ \text{moles of N}_2 = \frac{\text{moles of Al}}{2} = \frac{0.6856931060044478 \text{ mol}}{2} = 0.3428465530022239 \text{ mol} \][/tex]
3. Calculate the volume of [tex]\( \text{N}_2 \)[/tex] gas using the Ideal Gas Law:
The ideal gas law is given by:
[tex]\[ PV = nRT \][/tex]
rearranged to solve for volume ([tex]\( V \)[/tex]):
[tex]\[ V = \frac{nRT}{P} \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure in atmospheres (atm),
- Given pressure is 892 torr.
- Convert torr to atm: [tex]\( 1 \text{ atm} = 760 \text{ torr} \)[/tex]
[tex]\[ \text{Pressure in atm} = \frac{892 \text{ torr}}{760 \text{ torr/atm}} = 1.1736842105263158 \text{ atm} \][/tex]
- [tex]\( n \)[/tex] is the number of moles of gas ([tex]\( 0.3428465530022239 \text{ mol} \)[/tex]),
- [tex]\( R \)[/tex] is the ideal gas constant ([tex]\( 0.0821 \text{ L.atm/(mol.K)} \)[/tex]),
- [tex]\( T \)[/tex] is the temperature in Kelvin,
- Given temperature is [tex]\( 95^\circ \text{C} \)[/tex],
- Convert to Kelvin: [tex]\( T = 95 + 273.15 = 368.15 \text{ K} \)[/tex].
Plugging in these values:
[tex]\[ V = \frac{(0.3428465530022239 \text{ mol}) (0.0821 \text{ L.atm/(mol.K)}) (368.15 \text{ K}) }{1.1736842105263158 \text{ atm}} \][/tex]
Calculating this:
[tex]\[ V = 8.829101046864142 \text{ Liters} \][/tex]
So, [tex]\( 8.829 \)[/tex] liters of [tex]\( \text{N}_2 \)[/tex] gas, measured at 892 torr and [tex]\( 95^\circ \text{C} \)[/tex], are required to completely react with 18.5 grams of aluminum.
1. Determine the moles of aluminum (Al):
- Given mass of aluminum ([tex]\( \text{Al} \)[/tex]) is 18.5 grams.
- The molar mass of aluminum ([tex]\( \text{Al} \)[/tex]) is 26.98 grams per mole.
We find the number of moles of aluminum using the formula:
[tex]\[ \text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} \][/tex]
[tex]\[ \text{moles of Al} = \frac{18.5 \text{ g}}{26.98 \text{ g/mol}} = 0.6856931060044478 \text{ mol} \][/tex]
2. Determine the moles of nitrogen ([tex]\( \text{N}_2 \)[/tex]) required:
- According to the balanced chemical reaction:
[tex]\[ 2 \text{Al} + \text{N}_2 \rightarrow 2 \text{AlN} \][/tex]
2 moles of aluminum ([tex]\( \text{Al} \)[/tex]) react with 1 mole of nitrogen ([tex]\( \text{N}_2 \)[/tex]).
Therefore, the moles of [tex]\( \text{N}_2 \)[/tex] required will be half the moles of [tex]\( \text{Al} \)[/tex]:
[tex]\[ \text{moles of N}_2 = \frac{\text{moles of Al}}{2} = \frac{0.6856931060044478 \text{ mol}}{2} = 0.3428465530022239 \text{ mol} \][/tex]
3. Calculate the volume of [tex]\( \text{N}_2 \)[/tex] gas using the Ideal Gas Law:
The ideal gas law is given by:
[tex]\[ PV = nRT \][/tex]
rearranged to solve for volume ([tex]\( V \)[/tex]):
[tex]\[ V = \frac{nRT}{P} \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure in atmospheres (atm),
- Given pressure is 892 torr.
- Convert torr to atm: [tex]\( 1 \text{ atm} = 760 \text{ torr} \)[/tex]
[tex]\[ \text{Pressure in atm} = \frac{892 \text{ torr}}{760 \text{ torr/atm}} = 1.1736842105263158 \text{ atm} \][/tex]
- [tex]\( n \)[/tex] is the number of moles of gas ([tex]\( 0.3428465530022239 \text{ mol} \)[/tex]),
- [tex]\( R \)[/tex] is the ideal gas constant ([tex]\( 0.0821 \text{ L.atm/(mol.K)} \)[/tex]),
- [tex]\( T \)[/tex] is the temperature in Kelvin,
- Given temperature is [tex]\( 95^\circ \text{C} \)[/tex],
- Convert to Kelvin: [tex]\( T = 95 + 273.15 = 368.15 \text{ K} \)[/tex].
Plugging in these values:
[tex]\[ V = \frac{(0.3428465530022239 \text{ mol}) (0.0821 \text{ L.atm/(mol.K)}) (368.15 \text{ K}) }{1.1736842105263158 \text{ atm}} \][/tex]
Calculating this:
[tex]\[ V = 8.829101046864142 \text{ Liters} \][/tex]
So, [tex]\( 8.829 \)[/tex] liters of [tex]\( \text{N}_2 \)[/tex] gas, measured at 892 torr and [tex]\( 95^\circ \text{C} \)[/tex], are required to completely react with 18.5 grams of aluminum.
Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. Your questions find answers at IDNLearn.com. Thanks for visiting, and come back for more accurate and reliable solutions.
