Question

LID

Question 3
20 pts
How many atoms of aluminum can be produced by the decomposition of 31.5 g of
Al2O3?
(Hint: Write and balance the equation first.)
09.30 x 10 22 Al atoms
1.86 x 1023 Al atoms
O 7.44 x 1023 Al atoms
O none of these
O 3.72 x 1023 Al atoms



I need help

Answer 1

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Step-by-step explanation:

2Al2O3 = 4Al + 3O2 (Decomposition)

31.5/101.961g = 0.3089mol

(0.3089mol/2) * 4 = 0.61788 mole of Al

0.61788 mole * 6.022x10^23 atoms/mol = Your answer