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

Question

Chemistry

Araceli Watson

31 July, 16:04

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

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Lilyana Juarez · 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