A 4.78-g sample of aluminum completely reacts with oxygen to form 6.67 g of aluminum oxide. What is the mass percent composition of aluminum in aluminum oxide?

Question

17661 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-30T22:44:14

$71.7 %$ $71.7%$

The percent composition of aluminium in the

resulting oxide will be the ratio between the mass of aluminijm and the total mass of the oxide, multiplied

by $100$ $100$ .

$% Al = \frac{m_{Al}}{m_{oxide}} \times 100$ $"% Al" = m_"Al"/m_"oxide" xx 100$

In your case, you know that a $4.78-g$ $"4.78-g"$ sample of aluminium reacts completely with excess oxygen to form an

unknown aluminium oxide.

Moreover, you know that this oxide has a mass of $6.67 g$ $"6.67 g"$ .

Since all the luminium that initially reacted with the oxygen is now a part of the oxide, you can say that its

percent composition in the oxide will be

$"% Al" = (4.78color(red)(cancel(color(black)("g"))))/(6.67color(red)(cancel(color(black)("g")))) xx 100 = 71.664%$

Rounded to three sig figs, the

answer will be

$"% Al" = color(green)(71.7%)$