What is the maximum amount of zno (81.4084 g/mol) which could be formed from 12.85 g of zns (97.474 g/mol) and 15.91 g of o2 (31.9988 g/mol)?

             8.75 g of ZnO

Solution:

The Balance Chemical Reaction is as follow,

                                 2 ZnS  +   3 O₂    →    2 ZnO  +  2 SO₂  

Step 1: Find out the limiting reagent as;

According to Equation ,

            194.94 g (2 mole) ZnS reacts with  =  96 g (3 moles) of O₂

So,

                  12.85 g of ZnS will react with  =  X g of O₂

Solving for X,

                      X  =  (12.85 g × 96 g) ÷ 194.94 g

                      X  =  6.32 g of O₂

It means for total utilization of 12.85 g of ZnS we require 6.32 g of O₂, but we are provided with 15.91 g of O₂. Therefore, ZnS is the limiting reagent and will control the yield.

Step 2: Calculate Amount of ZnO produced as;

According to Equation ,

              194.94 g (2 mole) ZnS produces  =  132.81 g (2 moles) of ZnO

So,

                       12.85 g of ZnS will produce  =  X g of ZnO

Solving for X,

                      X =  (12.85 g × 132.81 g) ÷ 194.94 g

                      X  =  8.75 g of ZnO

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