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Zinc reacts with HCl to produce ZnCl2 and H2 gas. What is the volume (in mL) of 0.470 M HCl is required to react with 0.15 g of Zn metal?

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  • 8 months ago

    Zn + 2HCl ---> ZnCl2 + H2

    moles of Zn = 0.15/65.38 = 2.294*10^-3 mol

    moles of HCl = (2.294*10^-3)(2) = 4.588*10^-3 mol (from stoichiometry)

    volume of HCl = (4.588*10^-3)/0.470 = 9.76*10^-3 L = 9.76 mL

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