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?

Sep 8, 2017

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:

$n = \text{number of moles}$
$m = \text{mass}$
$M = \text{molar mass}$

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

$n = \frac{m}{M}$

The mass is given in the question:
${m}_{\text{Al}} = 14.5 g$
${m}_{\text{Oxygen gas}} = 7.5 g$

Molar mass of Al and Oxygen can be found in periodic table,
M_"Al"≈27g/mol
${M}_{\text{Oxygen gas"= 2*M_"Oxygen atom"≈2* 16=32 "g/mol}}$
(since oxygen gas (${O}_{2}$) has two atoms of Oxygen (O))

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

$n = \frac{m}{M} = \frac{14.5}{27} = 0.537 m o l s$

For ${O}_{2}$

$n = \frac{7.5}{32} = 0.234 m o l s$

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

For Al

$\frac{n}{4} = \frac{0.537}{4} = 0.134$

For ${O}_{2}$

$\frac{n}{2} = \frac{0.234}{2} = 0.17$

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

$\text{the value of Al (0.134)"< "the value of O_2 (0.17)}$

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. :)