how many liters of O2 gas measured at 784mmHg and 25'C are required to completely react with 2.3 mole of Al?
4Al + 3O2 ==> 2Al2O3
mols Al = 2.3 from the problem.
Convert to mols O2 required by using the coefficients in the balanced equation.
Then use PV = nRT and solve for volume in L.
To find the number of liters of O2 gas required to react with 2.3 moles of Al, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, we need to convert the pressure from mmHg to atm and the temperature from Celsius to Kelvin.
1 atm = 760 mmHg (approximately)
In this case, the pressure is 784 mmHg. Thus,
P = 784 mmHg / 760 mmHg/atm = 1.03 atm
The temperature is given as 25°C. To convert it to Kelvin, we add 273.15 to the Celsius temperature.
T = 25°C + 273.15 = 298.15 K
Substituting these values into our equation, we have:
PV = nRT
(1.03 atm) V = (2.3 mol) (0.0821 L·atm/(mol·K)) (298.15 K)
Now, we can solve for V:
V = (2.3 mol) (0.0821 L·atm/(mol·K)) (298.15 K) / (1.03 atm)
Calculating, we get:
V ≈ 56.44 Liters
Therefore, approximately 56.44 liters of O2 gas measured at 784 mmHg and 25°C are required to completely react with 2.3 moles of Al.