How many grams of zinc chloride could be formed from the reaction of 2.957 g of zinc with excess HCl?

Answer 1

Drop the e-2 from the answer you've been given! If you need greater accuracy (which I would expect, seeing as you've been given a quantity to 3 decimal places, use more accurate figures for the mass of Zn and ZnCl2 than 66 and 137.
We are assuming you follow the reasoning behind the equations you've been given? If not, let us know.

Answer 2

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THE ANSWER
I start from giving you accurate values of Atomic and Molecular Mass as

ZINC 65.37

ZINC CHLORIDE 136.27

Well, a mass like 2.957 g of ZINC are related to 45.2E-3 mol of the same metal

2.957 / 65.37 = 4.5E-2 mol

Stoichiometrically speaking, ANY MOLE of ZINC REACTS FORMING ONE MOLE of ZINC CHLORIDE.
45.2E-3 mol of chloride is like 6.16 g

4.5E-2 * 136.27 = 6.2 g.


I hope this helps you.

Answer 3

Zn + 2HCl -----> ZnCl2 + H2 5.206 g Zn (a million mol / sixty 5.38 g Zn) = 0.0796 mol Zn (a million mol ZnCl2 / a million mol Zn) = 0.0796 mol ZnCl2 (136.315 g ZnCl2 / a million mol) = 10.80 5 g ZnCl2 produced.