Question

Question #0601d

Answer 1

`"197 g"`

Explanation:

If I understand your question correctly, the answer will be `"197 g"`.

The trick here is to realize that the combined mass of the reactants must be equal to the combined mass of the products `->` think the Law of conservation of mass here.

You know that you're mixing `"20 g"` of aluminium metal and `"200 g"` of bromine, but that the reaction only consumes

`overbrace("200 g")^(color(blue)("what is available")) - overbrace("23 g")^(color(blue)("what is left behind")) = "177 g"`

of bromine. This means that the combined mass of the reactants will be

`"20 g + 177 g = 197 g"`

Since the reaction will only produce aluminium bromide, you can say that the mass of the product must be equal to `"197 g"`.

CHECK THE RESULT USING MOLES

If you want, you can check the result by using the balanced chemical equation

`2"Al"_ ((s)) + 3"Br"_ (2(l)) -> "Al"_ 2"Br"_ (6(aq))`

Use the molar masses of the two reactants to calculate how many moles of each take part in the reaction--keep in mind that only `"177 g"` of bromine react!

`20 color(red)(cancel(color(black)("g"))) * "1 mole Al"/(27.0color(red)(cancel(color(black)("g")))) = "0.74 moles Al"`

`177 color(red)(cancel(color(black)("g"))) * "1 mole Br"_2/(159.8color(red)(cancel(color(black)("g")))) = "1.11 moles Br"_2`

Aluminium reacts with bromine in a `2:3` mole ratio

`0.74 color(red)(cancel(color(black)("moles Al"))) * "3 moles Br"_2/(2color(red)(cancel(color(black)("moles Al")))) = "1.11 moles Al"`

and produces aluminium bromide in a `2:1` mole ratio

`0.74 color(red)(cancel(color(black)("moles Al"))) * ("1 mole Al"_2"Br"_6)/(2color(red)(cancel(color(black)("moles Al")))) = "0.37 moles Al"_2"Br"_6`

Finally, to convert this to grams, use the molar mass of aluminium bromide

`0.37 color(red)(cancel(color(black)("moles Al"_2"Br"_6))) * "533.4 g"/(1color(red)(cancel(color(black)("mole Al"_2"Br"_6)))) ~~ "197 g"`

Without taking into account the number of sig figs you have for your values, you can say that th answer will once again be `"197 g"` of aluminium bromide.