Question

Zinc and sulphur react to zinc sulphide if 25.0g of zinc and 30.0 g of sulphur are mixed.

A. Which chemical is the limiting reactant?
B. How many grams of ZnS will be formed?
C. How many grams of the excess reactant will remain after the reaction is over? ​

Answer 1

of zinc sulfide:

0.382 moles ZnS * 97.45 g/mol = 37.4 g ZnS

To find the mass of the excess reactant (sulfur) that will remain after the reaction is complete, we need to subtract the mass of the product (zinc sulfide) from the mass of the excess reactant:

30.0 g sulfur - 37.4 g ZnS = -7.4 g excess sulfur

Since the mass of the excess reactant is negative, this means that the reaction will consume all of the sulfur and there will be none left over.

To summarize:

A. Zinc is the limiting reactant.

B. 37.4 g of ZnS will be formed.

C. There will be no excess sulfur left over after the reaction is complete.

Step-by-step explanation: