If #50*g# of aluminum is oxidized, what mass of alumina will be recovered?

1 Answer

Answer:

Approx. #95*g# of alumina will be recovered.

Explanation:

We follow the stoichiometric equation........

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

And this 2 equiv of metal required 3/2 equivalents of dioxygen gas.

#"Moles of aluminum"# #=# #(50*g)/(27.0*g*mol^-1)=1.85*mol#.

And since aluminium is CLEARLY the so-called limiting reagent in this reaction (since we know precisely how much mass of #Al# reacted), at most we can get half an equiv of alumina....

And thus the mass of alumina is given by the product....

#1.85*molxx1/2xx101.96*g*mol^-1=??*g#.

Look for videos on molar equivalence and stoichiometry on the web. Perhaps this one: