# Question 0601d

Jul 20, 2017

$\text{197 g}$

#### Explanation:

If I understand your question correctly, the answer will be $\text{197 g}$.

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

You know that you're mixing $\text{20 g}$ of aluminium metal and $\text{200 g}$ 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

$\text{20 g + 177 g = 197 g}$

Since the reaction will only produce aluminium bromide, you can say that the mass of the product must be equal to $\text{197 g}$.

CHECK THE RESULT USING MOLES

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

$2 {\text{Al"_ ((s)) + 3"Br"_ (2(l)) -> "Al"_ 2"Br}}_{6 \left(a q\right)}$

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 $\text{177 g}$ 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 $\text{197 g}$ of aluminium bromide.