if 160g of Al reacts with O2 to produce 260g of Al2O3.calculate the theoretical yield and the percent yield
4Al+3O2 2Al2 O3
1. You have the balanced equation.
2. Convert 160 g Al to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles Al to moles Al2O3.
4. Now convert moles Al to grams. g = moles x molar mass. This is the theoretical yield in grams.
5. Percent yield = (actual yield/theoretical yield)*100 = ??
The problem states that 260 g is the actual yield.
To calculate the theoretical yield and percent yield, we need to follow these steps:
Step 1: Determine the molar masses of Al and Al2O3.
- The molar mass of Al (aluminum) is 26.98 g/mol.
- The molar mass of Al2O3 (aluminum oxide) is 101.96 g/mol.
Step 2: Calculate the number of moles of Al used.
- Moles of Al = mass of Al / molar mass of Al
= 160 g / 26.98 g/mol
≈ 5.93 mol
Step 3: Use the balanced equation to determine the moles of Al2O3 formed.
- According to the balanced equation: 4Al + 3O2 -> 2Al2O3
- From the equation, we can see that 4 moles of Al react to produce 2 moles of Al2O3.
- So, the moles of Al2O3 formed = (5.93 mol Al * 2 mol Al2O3) / 4 mol Al
≈ 2.96 mol
Step 4: Calculate the theoretical yield of Al2O3.
- Theoretical yield = moles of Al2O3 * molar mass of Al2O3
= 2.96 mol * 101.96 g/mol
≈ 302.12 g
Therefore, the theoretical yield of Al2O3 is approximately 302.12 grams.
Step 5: Calculate the percent yield.
- Percent yield = (actual yield / theoretical yield) * 100
= (260 g / 302.12 g) * 100
≈ 86%
Therefore, the percent yield of the reaction is approximately 86%.
To calculate the theoretical yield and percent yield, we need to understand a few concepts and use some formulas.
1. Balanced Equation: The given balanced equation is 4Al + 3O2 -> 2Al2O3. This tells us that for every 4 moles of aluminum (Al) reacting with 3 moles of oxygen (O2), we get 2 moles of aluminum oxide (Al2O3).
2. Molar Mass: The molar mass of aluminum (Al) is 26.98 g/mol, and the molar mass of aluminum oxide (Al2O3) is 101.96 g/mol.
Now let's calculate the theoretical yield:
Step 1: Convert the mass of aluminum (Al) to moles.
The molar mass of Al is 26.98 g/mol.
Number of moles of Al = Mass of Al / Molar mass of Al
Number of moles of Al = 160 g / 26.98 g/mol
Number of moles of Al = 5.93 mol (rounded to two decimal places)
Step 2: Calculate the moles of aluminum oxide (Al2O3) using the mole ratio from the balanced equation.
From the balanced equation, we know that 4 moles of Al will produce 2 moles of Al2O3.
Moles of Al2O3 = (Number of moles of Al / 4) * 2
Moles of Al2O3 = (5.93 mol / 4) * 2
Moles of Al2O3 = 2.96 mol (rounded to two decimal places)
Step 3: Convert the moles of aluminum oxide (Al2O3) to the mass.
Mass of Al2O3 = Moles of Al2O3 * Molar mass of Al2O3
Mass of Al2O3 = 2.96 mol * 101.96 g/mol
Mass of Al2O3 = 301.88 g (rounded to two decimal places)
The theoretical yield of aluminum oxide (Al2O3) is 301.88 g.
Now let's calculate the percent yield:
The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100%.
Percent yield = (Actual yield / Theoretical yield) * 100%
In this case, the actual yield is given as 260 g, and the theoretical yield is calculated as 301.88 g.
Percent yield = (260 g / 301.88 g) * 100%
Percent yield = 86.00% (rounded to two decimal places)
The percent yield of the reaction is 86.00%.