Answer : The amount of oxygen gas collected are, 0.217 mol
Explanation :
Using ideal gas equation :
[tex]PV=nRT[/tex]
where,
P = pressure of gas = [tex](790-12)torr=778torr=1.02atm[/tex] (1 atm = 760 torr)
V = volume of gas = 5 L
T = temperature of gas = [tex]14^oC=273+14=287K[/tex]
n = number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
[tex](1.02atm)\times (5L)=n\times (0.0821L.atm/mol.K)\times (287K)[/tex]
[tex]n=0.217mole[/tex]
Thus, the amount of oxygen gas collected are, 0.217 mol