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Question #ae3df

1 Answer

Apr 20, 2017

The reaction should be #4Al+3O_2->2Al_2O_3#

Explanation:

The molar ratio between #Al# and #O_2# is #4div3#

#Al(mol)div4.50(mol)=4div3# or #(Al)/4.50=4/3->Al=6(mol)#

Check:
The molar ratio between #Al# and #Al_2O_3# is #2div1#

So #6mol# of #Al# will give #6div2=3mol# of #Al_2O_3#

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