Oxygen gas reacts with powdered aluminum according to the following reaction:
4Al(s)+3 O2(g) --> 2 Al2O3 (s)
How many liters of O2 gas, measured at 789 mm Hg and 35 C, are required to completely react with 2.2 mol of Al?
for every 4 mols of Al you need 3 mols of O2
so
mols O2 = (3/4)mols Al = .75 *2.2 = 1.65 mols
Now do P V = n R T with n = 1.65 mol
To solve this stoichiometry problem, we need to follow these steps:
Step 1: Convert 2.2 mol of Al to mol of O2 using the mole ratio from the balanced equation.
From the balanced equation, we can see that the mole ratio between Al and O2 is 4:3. This means that for every 4 moles of Al, we need 3 moles of O2.
Therefore, we can set up the following ratio:
(2.2 mol Al) x (3 mol O2 / 4 mol Al) = 1.65 mol O2
Step 2: Convert the mole of O2 to liters of O2 using the ideal gas law.
The ideal gas law is expressed as 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 789 mm Hg to atm:
789 mm Hg x (1 atm / 760 mm Hg) = 1.04 atm
Next, we need to convert the temperature from Celsius to Kelvin:
35°C + 273.15 = 308.15 K
Now, we can use the ideal gas law to calculate the volume of O2:
V = (nRT) / P
V = (1.65 mol) x (0.0821 L.atm/mol.K) x (308.15 K) / (1.04 atm)
V ≈ 39.4 L
Therefore, approximately 39.4 liters of O2 gas, measured at 789 mm Hg and 35°C, are required to completely react with 2.2 mol of Al.
To determine the volume of O2 gas required to react with 2.2 mol of Al, we need to use the ideal gas law equation: PV = nRT.
First, let's list the given information:
- Pressure (P) = 789 mm Hg
- Temperature (T) = 35°C (308 K)
- Number of moles of Al (n) = 2.2 mol
- Balanced equation: 4Al(s) + 3O2(g) → 2Al2O3(s)
From the balanced equation, we can see that the stoichiometric ratio of Al to O2 is 4:3. This means that for every 4 moles of Al, we need 3 moles of O2.
Now, let's calculate the number of moles of O2 required:
- For the given number of moles of Al (2.2 mol), we can set up the following proportion to find the number of moles of O2:
4 mol Al / 3 mol O2 = 2.2 mol Al / x mol O2
Cross-multiplying, we get:
4 mol Al * x mol O2 = 2.2 mol Al * 3 mol O2
4x = 6.6
x = 6.6 / 4
x = 1.65 mol O2
Now, we can use the ideal gas law to determine the volume of O2 gas in liters:
PV = nRT
Where:
- P = Pressure in atmospheres (convert mm Hg to atm: 1 atm = 760 mm Hg)
- V = Volume in liters
- n = Number of moles
- R = Gas constant (0.0821 L·atm/mol·K)
- T = Temperature in Kelvin
Plug in the values:
P = 789 mm Hg = 789/760 atm
T = 35°C = 35 + 273 = 308 K
n = 1.65 mol
R = 0.0821 L·atm/mol·K
Now, substitute the values into the ideal gas law equation:
(789/760) atm * V = 1.65 mol * (0.0821 L·atm/mol·K) * 308 K
Simplifying:
V = (1.65 * 0.0821 * 308) / (789/760)
V = 0.128 L
Therefore, approximately 0.128 liters of O2 gas, measured at 789 mm Hg and 35°C, are required to completely react with 2.2 mol of Al.