Toy sparklers often contain aluminum powder, which burns in oxygen (O 2 ) to pro

Question

Toy sparklers often contain aluminum powder, which burns in oxygen (O2) to produce aluminum oxide and gives off a bright white light. The balanced reaction is:

4Al(s) + 3O2(g) ? 2Al2O3(s)

If only 2 moles of Al are available to react with abundant oxygen, Blank 1 mole(s) of aluminum oxide will be produced.

If 3 moles of aluminum oxide are produced, Blank 2 mole(s) of aluminum reacted.

If 3 moles of aluminum oxide are produced, Blank 3 moles (1 dec place) of oxygen reacted.

If 25.491 g of aluminum oxide is produced, Blank 4 g (3 dec places) of aluminum reacted?

Explanation / Answer

Given sparklers burn in the atmosphere there is always going to be an excess of oxygen.

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

(a) The balanced equation shows 4 mol of Al ---> 2 mol of Al2O3
So 2 mol of Al will produce 1 mol of Al2O3

(b) 1 mol of Al2O3 is formed by 2 mol Al. So 3 mol of Al2O3 will be formed from 6 mol of Al

(c) 2 mol of Al2O3 are formed from 3 mol of O2
So 1 mol is formed from 3/2 mol of O2
So 3 mol are formed from 9/2 = 4.5 mol of O2.

(d) Mm(Al2O3) = 102 g mol^-1
So 25.491 g contains 25.491/102 mol = 0.25 mol
1 mol of Al2O3 is formed from 2 mol of Al
So 0.25 mol of Al2O3 are formed from 0.5 mol of Al
Mm(Al) = 27 g mol^-1. So mass of 0.5 mol = 13.5 g

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