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

1 answer:

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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