Oxygen gas reacts with powdered aluminum according to the following reaction:
4Al(s)+3O2(g) yields 2Al2O3(s)
What volume of gas (in L), measured at 770 mmHg and 35 degrees C, is required to completely react with 53.9g of Al?
1. Convert 53.9 g Al to moles.
2. Using the coefficients in the balanced equation, convert moles Al to moles O2.
3. Use PV = nRT to convert moles O2 to volume at the conditions listed.
To find the volume of gas needed, we need to follow these steps:
Step 1: Convert grams of aluminum (Al) to moles.
To do this, we need the molar mass of aluminum (Al), which is 26.98 g/mol.
Divide the given mass of aluminum (53.9 g) by the molar mass to find the number of moles:
53.9 g Al / 26.98 g/mol = 1.998 moles Al
Step 2: Use the stoichiometry of the balanced equation to find the moles of oxygen (O2) required.
From the balanced equation, we can see that:
4 moles Al reacts with 3 moles O2
Therefore, the number of moles of O2 required is: (1.998 moles Al) x (3 moles O2 / 4 moles Al) = 1.4985 moles O2
Step 3: Convert moles of oxygen (O2) to volume using the ideal gas law.
The ideal gas law is represented as PV = nRT, where:
P = pressure in atm (converted from mmHg by dividing by 760)
V = volume in liters (what we need to find)
n = number of moles of gas (1.4985 mol O2)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (35 degrees C + 273.15 K = 308.15 K)
By rearranging the formula, we can solve for the volume:
V = (nRT) / P
V = (1.4985 mol) * (0.0821 L·atm/mol·K) * (308.15 K) / (770 mmHg / 760 mmHg/atm)
V ≈ 0.949 L
So, the volume of gas required to completely react with 53.9g of Al is approximately 0.949 L when measured at 770 mmHg and 35 degrees C.
To find the volume of gas required to completely react with 53.9g 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 calculate the number of moles of aluminum (Al) using its molar mass:
Molar mass of Al = 26.98 g/mol
Number of moles of Al = mass of Al / molar mass of Al
Number of moles of Al = 53.9 g / 26.98 g/mol
Number of moles of Al ≈ 2.00 mol (rounded to two decimal places)
From the balanced chemical equation, we can see that 4 moles of Al react with 3 moles of O2. Therefore, the number of moles of O2 can be calculated as follows:
Number of moles of O2 = (Number of moles of Al / 4) * (3 / 2)
Number of moles of O2 ≈ 1.50 mol (rounded to two decimal places)
Now, we can substitute the values into the ideal gas law equation to find the volume:
PV = nRT
V = (nRT) / P
V = (1.50 mol * 0.0821 L·atm/(mol·K) * (273.15 + 35) K) / (770 mmHg * 1 atm / 760 mmHg)
V ≈ 1.323 L (rounded to three decimal places)
Therefore, approximately 1.323 liters of gas, measured at 770 mmHg and 35 degrees Celsius, is required to completely react with 53.9g of Al.