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bagirrra123 [75]
3 years ago
12

Many grams of aluminum are required to produce 3.5 moles Al2O3 in the presence of excess O2?

Chemistry
1 answer:
Aleks [24]3 years ago
4 0
The  grams  of aluminum  that are required   to produce  3.5  moles of AlO3  in  presence of excess O2   is calculated as  below

write  the  equation for reaction
4 Al + 3O2 =2 Al2O3

by use of mole  ratio between  Al  to  Al2O3   which  is  4 :2  the moles of  Al 
=3.5 x4/2 = 7  moles

mass of Al  =  moles /   x molar mass

= 7 moles  x27 g/mol  =189  grams

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