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.
