Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g)   How many milliliters of 2.00 M HCl(aq) are required to react with 4.85 g Zn(s)?  volume:?

Please help! I am having trouble with this problem 

1 Answer

Relevance
  • 4 weeks ago

    Molar mass of Zn = 65.4 g/mol

    Balance equation for the reaction:

    Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

    Mole ratio Zn : HCl = 1 : 2

    Moles of Zn reacted = (4.85 g) / (65.4 g/mol) = 0.07416 mol

    Moles of HCl required = (0.07416 mol) × 2 = 0.1483 mol

    Volume of HCl required = (0.1483 mol) / (2.00 mol/L) = 0.0742 L = 74.2 mL

    ====

    OR:

    (4.85 g Zn) × (1 mol Zn / 65.4 g Zn) × (2 mol HCl / 1 mol HCl) × (1000 mL HCl / 2.00 mol HCl)

    = 74.2 mL HCl

Still have questions? Get your answers by asking now.