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#?

1 Answer
Sep 8, 2017

Answer:

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. :)