Answer to Question #274107 in General Chemistry for FutureBSP

Question

1. 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 grams of ZnS is produced when zinc is completely used up?

B. Which reactant is the limiting reagent?

C. How many grams of the excess reagent are left?

Answer 1

MM(Zn) = 65.37 g/mol

MM(S) = 32.06 g/mol

MM(ZnS) = 97.43 g/mol

$n(Zn) = \frac{36.8}{68.37} = 0.538 \; mol \\ n(S) = \frac{19.4}{32.06} = 0.605 \; mol$

A. n(ZnS) =n(Zn) = 0.538 mol

$m(ZnS) = 0.538 \times 97.53 = 52.41 \; g$

B. n(Zn) < n(S) => Zn is a limitting reactant.

C. Used n(S) = n(Zn) = 0.538

Remaining n(S) = 0.605 – 0.538 = 0.067 mol

Remaining $m(S) = 0.067 \times 32.06 = 2.15 \;g$