Question

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

Answer 1

First you have to find how many moles of Al2O3 are in 25.0 g. To find that you divide the amount of grams by the molar mass.
molar mass = 2(26.98) + 2(16) =85.96
Now to find the number of moles of Al2O3 you have...
25 g / (85.96g /mol) = .291 moles (the grams cancelled out)

Now that ratio of the amount of Aluminum to Al2O3 is 2:1 (2 moles of Al for 1 mole of Al2O3)
so you much multiply the number of moles of Al2O3 by 2.
.291 (2) = .582 moles of Al

now Convert back to grams by multiplying the number of moles of Al by the molar mass of Al
so
.582 mol x 26.98 = 15.69 g

Answer 2

`13.22g Al`

Step-by-step explanation:

First we have to find the molar mass of the compound.

We must multiply the atomic mass of each element of the compound by the number of atoms of each one

Aluminum oxide:
`Al_(2)O_(3)`

molar mass
`Al_(2)O_(3)`

`Al= 26.98g/mol\\O= 16g/mol`

`Al_(2)O_(3)= 2(26,98g/mol)+3(16g/mol)=101.96g/mol`

We know that in 1 mol of
`Al_(2)O_(3)` it has a mass of 101.96g.

Now we can find out how many moles of
`Al_(2)O_(3)` are in 25 g.

`101.96 g Al_2O_3\longrightarrow 1 mol Al_2O_3\\ 25g Al_2O_3\longrightarrow x\\ x=(25g Al_2O_3.1 molAl_2O_3)/(101.96 g Al_2O_3) =0.245molAl_2O_3`

For every mole of
`Al_(2)O_(3)` we have 2 moles of Al.

To find out how many moles of Al we have in
`0.245molAL_2O_3` we use a rule of three

`1 mol Al_2O_3\longrightarrow 2mol Al \\ 0.245mol Al_2O_3\longrightarrow x\\ x=(0.245mol Al_2O_3. 2 molAl)/(1 mol Al_2O_3) =0.49molAl`

There are 0.49 moles of Al in 25 g of
`Al_(2)O_(3)`

now we calculate the grams of Al in 0.49 moles of Al

`0.49 mol Al.(26.98g Al)/(1 mol de Al)= 13.22 g Al`