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