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

idk what the question is about

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.