# Question f373b

Jul 6, 2017

${\text{0.81 g Al"_2"S}}_{3}$

#### Explanation:

The balanced chemical equation that describes this reaction looks like this

$2 {\text{Al"_ ((s)) + 3"S"_ ((s)) -> "Al"_ 2"S}}_{3 \left(s\right)}$

Notice that you need $2$ moles of aluminium in order to produce $1$ mole of aluminium sulfide. Since sulfur is said to be in excess, you don't need to worry about this reactant, i.e. you have more than enough sulfur to ensure that all the sample of aluminium reacts.

Now, you know that aluminium has a molar mass of ${\text{26.98 g mol}}^{- 1}$, which means that $1$ mole of aluminium has a mass of $\text{26.98 g}$.

Aluminium sulfide has a molar mass of ${\text{150.16 g mol}}^{- 1}$, which, of course, implies that $1$ mole of aluminium sulfide has a mass of $\text{150.16 g}$.

You can thus say that for every

2 color(red)(cancel(color(black)("moles Al"))) * "26.98 g"/(1color(red)(cancel(color(black)("mole Al")))) = "53.96 g"

that take part in the reaction, you get

1 color(red)(cancel(color(black)("mole Al"_2"S"_3))) * "150.16 g"/(1color(red)(cancel(color(black)("mole Al"_2"S"_3)))) = "150.16 g"#

of aluminium sulfide. This means that your sample will produce

$0.29 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g Al"))) * ("150.16 g Al"_2"S"_3)/(53.96color(red)(cancel(color(black)("g Al")))) = color(darkgreen)(ul(color(black)("0.81 g}}}}$

of aluminium sulfide. The answer is rounded to two sig figs, the number of sig figs you have for the mass of aluminium.