How many moles of zncl2 will be produced from 61.0g of zn, assuming hcl is available in excess?

            127.15 g of ZnCl₂

Solution:

             The balance chemical equation is as follow,

                                  Zn  +  2 HCl    →    ZnCl₂  +  H₂

According to equation,

           65.38 g (1 mole) of Zn produced  =  136.28 g (1 mole) of ZnCl₂

So,

                       61.0 g of Zn will produce  =  X g of ZnCl₂

Solving for X,

                      X  =  (61.0 g × 136.28 g) ÷ 65.38 g

                      X  =  127.15 g of ZnCl₂