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