Zn+2hcl=zncl2+h2 how many moles of zncl2 will be produced from 61.0 g of zncl2, 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

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