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

Md Ariful Islam

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Md Ariful Islam

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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. :)

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