How many grams of aluminum are in 25.0 g of aluminum oxide?

Question

Peyton Adams · Accepted Answer

#"%aluminium by mass"# #=# #"Molar mass of aluminum"/"Molar mass of alumina"xx100%..# #"Approx 27%....."#

Thus, the percentage of metal in a molar amount of aluminum is,

#(26.98*g*mol^-1)/(101.96*g*mol^-1)xx100%=??#

And in this mass of alumina there are approx. #6.6*g# of metal.

Olivia Anton · Answer

undefined In order to calculate the mass of aluminum in aluminum oxide, you must first calculate the molar mass of aluminum oxide, which is $101.96 \, \text{g/mol} $. Then, divide the mass of aluminum (26.98 g/mol) by the molar mass of aluminum oxide, and multiply the result by the mass of aluminum oxide that has been given (25.0 g).
$$ \text{Mass of aluminum} = \left( \frac{26.98 \, \text{g/mol}}{101.96 \, \text{g/mol}} \right) \times 25.0 \, \text{g} $$
$$ \text{Mass of aluminum} = \frac{26.98}{101.96} \times 25.0 \, \text{g} $$
$$ \text{Mass of aluminum} = 6.665 \, \text{g} $$ As a result, there are roughly 6.67 grams of aluminum in 25.0 grams of aluminum oxide.

HIX Tutor · Answer

undefined To find out how many grams of aluminum are in 25.0 g of aluminum oxide, follow these steps:

1. **Learn the formula**: Aluminum oxide is Al₂O₃. This means it has 2 aluminum (Al) atoms and 3 oxygen (O) atoms.

2. **Find the molar mass**:
   - Aluminum (Al) = 27.0 g/mol (2 Al in Al₂O₃: 2 × 27.0 = 54.0 g)
   - Oxygen (O) = 16.0 g/mol (3 O in Al₂O₃: 3 × 16.0 = 48.0 g)
   - Total for Al₂O₃ = 54.0 g + 48.0 g = 102.0 g/mol

3. **Calculate the fraction of aluminum**: 
   - Aluminum mass in Al₂O₃ = 54.0 g
   - Total mass = 102.0 g
   - Fraction of aluminum = 54.0 g / 102.0 g = 0.5294 (approximately)

4. **Find grams of aluminum in 25.0 g of aluminum oxide**:
   - Multiply the fraction by the total weight: 
   - 25.0 g × 0.5294 = 13.2 g (approximately)

So, there are about **13.2 grams** of aluminum in 25.0 g of aluminum oxide.