Under certain conditions, aluminium metal can burn violently in the presence of oxygen to form aluminium oxide. The balanced equation for this reaction is shown below. What volume of oxygen gas at 1.00 atm and 298 K is needed for complete reaction with 0.880 mol Al?
4Al(s) + 3O2(g) → 2Al2O3(s)
From reaction,
Mol of O2 required = (3/4)*mol of Al
= (3/4)*0.880 mol
= 0.660 mol
Given:
P = 1.0 atm
n = 0.660 mol
T = 298.0 K
use:
P * V = n*R*T
1 atm * V = 0.66 mol* 0.08206 atm.L/mol.K * 298 K
V = 16.1 L
Answer: 16.1 L
Under certain conditions, aluminium metal can burn violently in the presence of oxygen to form aluminium...
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