# 2"Al"+3"S"→"Al"_2"S"_3 125g of Al2S3 was produced from 75g of Al & an excess of S What is the theoretical yield of Al2S3?

Jul 31, 2017

Well, the theoretical yield is $\text{210 g}$... what is the percent yield?

You don't really need to know that $\text{125 g}$ of ${\text{Al"_2"S}}_{3}$ were formed to answer this question. You know that $\text{75 g}$ of $\text{Al}$ were consumed, and it is obvious that sulfur is in excess (you were told that right in the question).

So, treat aluminum as your limiting reactant, the reactant whose consumption generates the maximum amount of product you can get.

$75 \cancel{\text{g Al" xx "1 mol"/(26.982 cancel"g Al") = "2.780 mols Al}}$

And this many mols of $\text{Al}$ will react with how much ${\text{Al"_2"S}}_{3}$? From the chemical reaction (which you should verify is balanced), the stoichiometry states:

${\text{2 mols Al" = "1 mol Al"_2"S}}_{3}$

So, this ratio is a unit conversion factor (i.e. a ratio that is equivalent to $1$ that gets you into a new way of expressing a given quantity):

"2 mols Al"/("1 mol Al"_2"S"_3)

And thus, we have...

2.780 cancel"mols Al" xx ("1 mol Al"_2"S"_3)/(2 cancel"mols Al")

$=$ $\text{1.390 mols product}$

And now we just use its molar mass to find its produced mass:

"1.390 mols Al"_2"S"_3 xx (2 xx "26.982 g" + 3 xx "32.065 g")/("1 mol Al"_2"S"_3)

$=$ $\text{208.69 g product}$

Unfortunately, we only get $\textcolor{b l u e}{\text{210 g}}$, because you've only given yourself two sig figs.