Question

Calculate the mass of alumina (Al2O3) produced when 100 g of aluminum burns in oxygen.

Answer 1

4Al + 3O2 -------> 2Al2O3

1)100 g Al into moles

M(Al) = 27.0 g/mol

100g Al* 1 mol Al/27 g Al = 100/27 g/mol Al

2) 4Al + 3O2 -------> 2Al2O3

from reaction 4 mol 2 mol

given 100/27 mol x mol

x= (100/27)*2/4=50/27 mol Al2O3

3) 50/27 mol Al2O3 into g

M(Al2O3) = 2*27.0 + 3*16.0= 102 g/mol

(50/27) mol Al2O3 * 102 g Al2O3/1 mol Al2O3 = (50*102)/27≈ 189 g Al2O3

Answer: 189 g/mol Al2O3.

Answer 2

The mass of alumina that would be produced when 100 g of aluminum is burnt in oxygen will be 188.95 g

Aluminum burns in oxygen according to the following equation:

                       $4Al + 3O_2 --> 2Al_2O_3$

This means that the mole ratio of aluminum burnt to that of alumina produced is 4:2 or 2:1.

Recall that: mole = mass/molar mass

mole of 100 g aluminum = 100/26.98

                                             = 3.71 moles

Thus, mole of alumina produced = 3.71 /2

                                                       = 1.85 moles

Mass of alumina produced = mole x molar mass

                                         = 1.85 x 101.96

                                          = 188.95 g

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