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How many milliliters of 4.80 M HCl are required to react with 110 mL of 1.48 M Al(OH)3?
Al(OH)3(s) + 3HCl(aq)to AlCl3(aq) + 3H2O(aq)

  • chem -

    mols Al(OH)3 = M x L = ?
    Convert mols Al(OH)3 to mols HCl.
    Then M HCl = mols HCl/L HCl. You know mols and M, solve for L and convert to mL.

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