Respuesta :
Answer:
11.9 L
Explanation:
Data Given:
Mass of Al₂O₃ = 44.18 g
Temperature = 290 K
Pressure = 1.3 atm
Volume = ?
Solution:
As aluminum metal react with Oxygen, therefore first look for the reaction
4Al + 3O₂ ----> 2Al₂O₃
So from here we will get information about moles of oxygen
First find number of moles of 44.18 grams of Al₂O₃
Molar mass of Al₂O₃ = 102 g/mol
convert mass to moles
no.of moles = mass in g / molar mass
put values in above formula
no.of moles = 44.18 g / 102 g/mol
no.of moles = 0.433 mol
Now as we have the following equation
4Al + 3O₂ ----------> 2Al₂O₃
3 mole 2 mole
So according to equation 2 mole Al₂O₃ form from 3 moles of oxygen the how many moles of oxygen will form 0.433 mole of Al₂O₃
3 moles of O₂ ≅ 2 moles of Al₂O₃
X moles of O₂ ≅ 0.433 moles of Al₂O₃
By cross multiplication
X moles of O₂ = 3 x 0.433 /2
moles of oxygen = 0.65 moles
Now find the Volume of Oxygen gas
for this formula used
PV = nRT
V = nRT / P. . . . . . . . . . (1)
where
n = no. of moles of O₂ = 0.65 mol
R = Gas Constant = 0.0821 L.atm.mol⁻¹.K⁻¹
Put values above formula (1)
V = 0.65 mol x 0.0821 L.atm.mol⁻¹.K⁻¹ x 290 K / 1.3 atm.
V = 11.9 L
So the volume of oxygen during experiment = 11.9 L
