Aluminum oxide forms when 3.50 moles of aluminum is combined with 3.00 moles of oxygen. What is the excess reagent? Al + O2 - ALO3
(unbalanced) *
A - oxygen
B - aluminum
C - aluminum oxide

Explanation:

Correct option is

B

18

As reaction for the formation of Al 2

is as follows:

4Al+3O

2

→Al 2

O

this shows that 4 mol reacts to 3 mol ( 96 g) of O

2

so as 3 mol O

2

to 4 mol Al

then 0.5 mol O 2 to 3, 4

×0.5=0.66 mol of Al

as 1 mol Al = 27 g

then 0.66 mol of Al=0.66×27=17.9 g≈18 g

Hence, the correct option is B.

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