# Question 86707

Jul 6, 2016

${\text{68 g Al"_2"O}}_{3}$

#### Explanation:

You need to write the balanced chemical equation that describes this synthesis reaction

$4 {\text{Al"_ ((s)) + color(blue)(3)"O"_ (2(g)) -> color(red)(2)"Al"_ 2"O}}_{3 \left(s\right)}$

Notice that the two reactants react in a $4 : \textcolor{b l u e}{3}$ mole ratio, which means that every $\textcolor{b l u e}{3}$ moles of oxygen gas that take part in the reaction require $4$ moles of aluminium metal.

Since you we told that aluminium is in excess, you can safely assume that all the oxygen will actually react.

Now, the reaction produces $\textcolor{red}{2}$ moles of aluminium oxide for every $\textcolor{b l u e}{3}$ moles of oxygen gas that take part in the reaction.

Since the given sample of oxygen gas reacts completely, you can say that the reaction will produce

1.0 color(red)(cancel(color(black)("moles O"_2))) * (color(red)(2)color(white)(a)"moles Al"_2"O"_3)/(color(blue)(3)color(red)(cancel(color(black)("moles O"_2)))) = 2/3color(white)(a)"moles Al"_2"O"_3

To convert the number of moles of aluminium oxide to grams, use the compound's molar mass

2/3 color(red)(cancel(color(black)("moles Al"_2"O"_3))) * "102 g"/(1color(red)(cancel(color(black)("mole Al"_2"O"_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("68 g Al"_2"O"_3)color(white)(a/a)|)))#