Asked by Josh Porter
If 12.5 g of Zinc reacts with excess hydrochloric aid, how many grams of zinc chloride are produced?
Answers
Answered by
Curry
So,
We have a balanced equation of:
Zn + 2HCL --> ZnCl2 + 2H
Zn is the limiting reactant.
g ZnCl2 = (12.5 g Zn) (1 mole Zn/65.39g) (1 mole ZnCl2/1 mole Zn) (136.29 g ZnCl2 /1 mol) = 26.05 or 26.1 grams, rounded to one sig fig.
We have a balanced equation of:
Zn + 2HCL --> ZnCl2 + 2H
Zn is the limiting reactant.
g ZnCl2 = (12.5 g Zn) (1 mole Zn/65.39g) (1 mole ZnCl2/1 mole Zn) (136.29 g ZnCl2 /1 mol) = 26.05 or 26.1 grams, rounded to one sig fig.
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