Answer to Question #155722 in General Chemistry for Willy

Question

In the production of Zinc sulfide, 36.8 g of zinc is made to react with 19.4 g of sulfur. 

𝑍𝑛 + 𝑆 → 𝑍𝑛𝑆

a. How many moles of ZnS is produced when sulfur is completely used up?

b. How many grams of ZnS is produced when zinc is completely used up?

c. Which reactant is the limiting reagent?

d. How many grams of the excess reagent is left?

Answer 1

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

8Zn + S₈ $\to$ 8ZnS

Noticed that you have 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 8:8 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

36.8g × $\frac{1molZn}{65.38g}$ = 0.562 moles Zn

For sulfur, you'll have

19.4g × $\frac{1molS_8}{256.52g}$ = 0.0756 moles S₈

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.

0.0756 moles S₈ × $\frac{8molesZn}{1moleS_8}$ = 0.6048 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

0.562 g Zn × $\frac{1molesS_8}{8moleZn}$ = 0.07025 moles S₈

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

0.562molesZn ×$\frac{8molesZnS}{8molesZn}$ = 0.562 moles ZnS

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

0.562molesZnS × $\frac{97.445g}{1molZnS}$ = 54.76g

Zinc is the limiting reagent. The reaction produces 54.76 g of zinc sulfide.

Excess reagent = 36.8 + 19.4 - 54.76

= 56.2 - 54.76 = 1.44g