(3 points) Let 124 grams of Al react with 601 grams of Fe2O3. What is the mass (in grams) of the formed Al2O3?

1 Answer
Jan 29, 2018

Approx. #230*g# of alumina result....

Explanation:

First we need a stoichiometrically balanced equation....

#Fe_2O_3(s) + 2Al(s) + Delta rarr 2Fe(s) + Al_2O_3(s)#

...and this is balanced with respect to mass and charge, as required.

And then we calculate equivalent quantities of metal and metal oxide....

#n_"ferric oxide"=(601*g)/(159.69*g*mol^-1)=3.76*mol#

#n_"aluminum"=(124*g)/(27.0*g*mol^-1)=4.59*mol#

And clearly, aluminum is the limiting reagent....do you agree...and at most half an equiv of ferric oxide can react according to the given equation....

#(4.59*mol)/2xx101.96*g*mol^-1=??*g#