# Given the equation: 8Zn +S_8 -> 8ZnS, if 7.36 g of zinc react with 6.45 g of sulfur, which is the limiting reactant and what mass of zinc sulfide forms?

Jan 10, 2016

Zinc is the limiting reagent. The reaction produces $\text{11.0 g}$ of zinc sulfide.

#### Explanation:

Take a look at the balanced chemical reaction for this synthesis reaction

$\textcolor{red}{8} {\text{Zn"_text((s]) + "S"_text(8(s]) -> color(purple)(8)"ZnS}}_{\textrm{\left(s\right]}}$

Notice that you have a $\textcolor{red}{8} : 1$ mole ratio between zinc and sulfur. This tells you that will always consume eight moles of zinc for every one mole of sulfur that takes part in the reaction.

Moreover, you have a $\textcolor{red}{8} : \textcolor{p u r p \le}{8}$ mole ratio between zinc and zinc sulfide, the product of the reaction. This tells you that the reaction will produce as many moles of zinc sulfide as you have moles of zinc that take part in the reaction.

Your strategy here will be to use the molar masses of the three chemical species that take part in the reaction to go from grams to moles for the reactants, and from moles to Grams for the product.

For zinc, you'll have

7.36 color(red)(cancel(color(black)("g"))) * "1 mole Zn"/(65.38color(red)(cancel(color(black)("g")))) = "0.11257 moles Zn"

For sulfur, you'll have

6.45 color(red)(cancel(color(black)("g"))) * "1 mole S"_8/(256.52color(red)(cancel(color(black)("g")))) = "0.025144 moles S"_8

Now, you need to use the mole ratio that exists between the two reactants to determine whether or not you have enough moles of zinc to react with that much moles of sulfur.

0.025144 color(red)(cancel(color(black)("moles S"_8))) * (color(red)(8)" moles Zn")/(1color(red)(cancel(color(black)("mole S"_8)))) = "0.20115 moles Zn"

Since you have fewer moles of zinc than the complete consumption of the sulfur would have required, you can say that zinc will act a a limiting reagent.

More specifically, only

0.11257 color(red)(cancel(color(black)("moles Zn"))) * "1 mole S"_8/(color(red)(8)color(red)(cancel(color(black)("moles Zn")))) = "0.014071 moles S"_8

will actually take part in the reaction, the rest will be in excess.

Now, if zinc is completely consumed by the reaction, it follows that you will end up with

0.11257color(red)(cancel(color(black)("moles Zn"))) * (color(purple)(8)" moles ZnS")/(color(red)(8)color(red)(cancel(color(black)("moles Zn")))) = "0.11257 moles ZnS"

Finally, use zinc sulfide's molar mass to determine how many grams would contain this many moles

0.11257 color(red)(cancel(color(black)("moles ZnS"))) * "97.445 g"/(1color(red)(cancel(color(black)("mole ZnS")))) = color(green)("11.0 g")

The answer is rounded to three sig figs.

Here's a quick video of how this reaction looks like