Question

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

Answer 1

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

Answer 2

Answer:

thanks... the volume of a mass is 189 g.

Explanation:

thank you for your help in need!

I made it on the USA test prep