Question

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

Answer 1

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