Question

A chemistry has 14.5 g of Al and 7.5 g of `O_2`. What would the excess be for the following reaction: `4Al + 3O_2 -> 2Al_2O_3`?

Answer 1

Al is in limiting
`O_2` is in excess

Explanation:

To check excess and limiting reactants, the number of moles has to be found, and then divided by the coefficient.
For the sake of shortening the equations, let us say that:

`n="number of moles"`
`m="mass"`
`M="molar mass"`

To find the number of moles, we need to use the formula:

`n=m/M`

The mass is given in the question:
`m_"Al"=14.5g`
`m_"Oxygen gas"=7.5g`

Molar mass of Al and Oxygen can be found in periodic table,
`M_"Al"≈27g/mol`
`M_"Oxygen gas"= 2*M_"Oxygen atom"≈2* 16=32 "g/mol"`
(since oxygen gas (`O_2`) has two atoms of Oxygen (O))

Now putting these values for the number of moles's formula:
For Al

`n=m/M=14.5/27=0.537 mols`

For `O_2`

`n=7.5/32=0.234 mols`

After getting the number of moles, we just still have to divide them by their respective coefficients, and comparing them.

For Al

`n/4=0.537/4=0.134`

For `O_2`

`n/2=0.234/2=0.17`

The reactant with the bigger value will be in excess and the reactant with the smaller value will be in limiting

`"the value of Al (0.134)"< "the value of O_2 (0.17)"`

Therefore, Al is in limiting (there is less of it and thus it is limiting the reaction) and `O_2` is in excess.

Hope that this will help. :)