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How many moles of ZnCl2 will be produced from 61.0g of Zn,assuming HCl is available in excess?

How many moles of ZnCl2 will be produced from 610g of Znassuming HCl is available in excess class=

Respuesta :

Answer:

            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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