How many liters of O2 gas, measured at 772 mmHg and 20 degrees C, are required to completely react with 2.4 mol of Al?
To determine the number of liters of O2 gas, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in this case, in mmHg)
V = Volume (we need to calculate this)
n = Number of moles of gas (in this case, for O2)
R = Gas constant (0.0821 L·atm/mol·K)
T = Temperature (in this case, in Kelvin)
First, let's convert the given pressure from mmHg to atm:
772 mmHg * (1 atm / 760 mmHg) = 1.0158 atm
Next, we need to convert the temperature from Celsius to Kelvin. The formula for this conversion is:
T(K) = T(°C) + 273.15
20 °C + 273.15 = 293.15 K
Now, we have all the values we need for the ideal gas law equation:
(1.0158 atm) * V = (n) * (0.0821 L·atm/mol·K) * (293.15 K)
Rearranging the equation to solve for V:
V = (n * R * T) / P
V = (2.4 mol * 0.0821 L·atm/mol·K * 293.15 K) / 1.0158 atm
Calculating this gives:
V ≈ 67.9 L
Therefore, approximately 67.9 liters of O2 gas are required to completely react with 2.4 mol of Al under the given conditions.