Question

Question #86707

Answer 1

`"68 g Al"_2"O"_3`

Explanation:

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

`4"Al"_ ((s)) + color(blue)(3)"O"_ (2(g)) -> color(red)(2)"Al"_ 2"O"_ (3(s))`

Notice that the two reactants react in a `4:color(blue)(3)` mole ratio, which means that every `color(blue)(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 `color(red)(2)` moles of aluminium oxide for every `color(blue)(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)|)))`