a piece of Aluminium weighing 2.7g is titrated with 75 mL of H2SO4(specific gravity 1.18 and 24.7% H2SO4 by weight). after the metal is completely dissolved the solution is diluted to 400 mL . calculate the molarity of free H2SO4 in the solution.

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  1. 2Al + 3H2SO4 ==> Al2(SO4)3 + 3H2

    mols H2SO4 initially = 1.18 g/mL x 75 mL x (24.7/100) x 1/98 = approx 0.228
    mols Al initially = 2.7g/27 = 0.1 mol

    All of the Al is dissolved. How much of the H2SO4 is used? That is
    0.1 mol Al x (3 mols H2SO4/2 mols Al) = 0.15 mols H2SO4 used.
    How much H2SO4 remains unreacted? That's 0.228 - 0.150 = ?
    M H2SO4 in final solution is mols/L = mols/0.400 L = ?
    Post your work if you get stuck.

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  2. I got the solution from you. I was stucked for hours. You saved my day. A heartful thanks to you!!! Keep growing.

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