Answer:
34.28 L ( 1.5*22.4 L)
Explanation:
Calculation of the moles of aluminum as:-
Mass = 55 g
Molar mass of aluminum = 26.981539 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{55\ g}{26.981539\ g/mol}[/tex]
[tex]Moles= 2.0384\ mol[/tex]
According to the reaction:-
[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]
4 moles of aluminum react with 3 moles of oxygen gas
1 mole of aluminum react with [tex]\frac{3}{4}[/tex] moles of oxygen gas
2.0384 moles of aluminum react with [tex]\frac{3}{4}\times 2.0384[/tex] moles of oxygen gas
Moles of oxygen gas = 1.5288 moles
At STP,
Pressure = 1 atm
Temperature = 273.15 K
Using ideal gas equation as:
[tex]PV=nRT[/tex]
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K
⇒V = 34.28 L ( 1.5*22.4 L)
