How many milliliters of 5.50 m hcl(aq) are required to react with 9.55 g of zn(s)?

Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g)  From the balanced chemical equation, 2 moles of HCl is required to react with 1 mole of Zn   Mass of Zn = 9.55 g  Molar mass of Zinc = 65.41 g/mol   Number of moles = mass / molar mass   Moles of Zn = 9.55 / 65.41   = 0.146 mol  Moles of HCl required = 2 x moles of Zn   = 2 x 0.146   = 0.292 mol

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