Zinc reacts with hydrochloric acid according to the reaction equation $Zn (s) + 2 HCl (aq) ⟶ ZnCl_{2} (aq) + H_{2} (g)$ How many milliliters of $5.50 M HCl (aq)$ are required to react with $3.05 g Zn (s)?$ volume:

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Zinc reacts with hydrochloric acid according to the reaction equation Zn(s)+2HCl(aq)ZnCl2(aq)+H2(g) How many milliliters of 5.50MHCl(aq) are required to react with 3.05gZn(s)? volume:

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