Question

How do we represent the oxidation of a `25*g` mass aluminum metal to alumina?

Answer 1

Well, we need a stoichiometric equation....

Explanation:

`Al(s) + 3/2O_2(g) rarr Al_2O_3(s)`

And so we interpolate the equivalent masses of aluminum and dioxygen....

`"Moles of aluminum"=(25.0*g)/(27.0*g*mol^-1)=0.926*mol`

...and this molar quantity of metal requires...

`3/2*"equiv"xx0.926*molxx32.00*g*mol^-1=44.5*g` with respect to dioxygen....and all I have done is to use the stoichiometric equation to inform me of the equivalent masses of metal and dioxygen....

Answer 2

47.309 grams

Explanation:

We need to write a balanced equation for this reaction.

`4Al + 3O_2 -> 2Al_2O_3`

Aluminium and aluminium oxide are in a 2:1 ratio.

You need to calculate the number of moles of aluminum in the 25g.

To do this you use the equation:

no. of moles (mol) = mass (g) / molecular weight (g/mol)

MW Al = 26.981 g/mol

no. of moles = 25 / 26.981

no. of moles = 0.927 mol (3 decimal places)

As there is a 2:1 ratio you must divide the number of moles by 2 as for every 2 moles of aluminium you produce 1 mole of aluminum oxide.

no. of moles of `Al_2O_3` = 0.927 / 2

no. of moles of `Al_2O_3` = 0.464 mol (3 decimal places)

To convert back to mass we use the same equation as before, but rearranged for mass.

mass = no. of moles X molecular weight

MW `Al_2O_3` = 101.96 g/mol

mass `Al_2O_3` = 0.464 X 101.96

mass `Al_2O_3` = 47.309 g