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    3 C part iii

    http://www.a-levelchemistry.co.uk/ol...une%202006.pdf

    I did the intial numbers of each moles take away each other (0.0144-0.0063 = 8.1x10-3) but the answer in the marck scheme is 1.8x10-3, apparently you times the 0.0063 by two but I don't see why?

    Cheers
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    (Original post by alevelsguy)
    3 C part iii

    http://www.a-levelchemistry.co.uk/ol...une%202006.pdf

    I did the intial numbers of each moles take away each other (0.0144-0.0063 = 8.1x10-3) but the answer in the marck scheme is 1.8x10-3, apparently you times the 0.0063 by two but I don't see why?

    Cheers
    Mol H2SO4 = 18.0cm3 of a 0.350 moldm = 0.0063 mol

    This reacts:

    H2SO4 + 2KOH --> products

    So mol of KOH reacted = 2 x 0.00603

    So mol of KOH remaining = init moles - 0.0126
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    (Original post by alevelsguy)
    3 C part iii

    http://www.a-levelchemistry.co.uk/ol...une%202006.pdf

    I did the intial numbers of each moles take away each other (0.0144-0.0063 = 8.1x10-3) but the answer in the marck scheme is 1.8x10-3, apparently you times the 0.0063 by two but I don't see why?

    Cheers
    First of all, write balanced equation. IMPORTANT!

    2KOH + H2SO4 ----> K2SO4 + 2H2O

    1) mol of KOH(total) = 0.0144 mol

    2) mol of acid reacted = 0.0063 mol

    3) from balanced eqn, mol of KOH reacted = 2x mol of acid = 0.0126 mol

    therefore mol of excess KOH= total - reacted = 0.0144 - 0.0126 = 1.8 x10^-3

    EDIT : You are faster, charco!
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    Where does it say that 2KOH reacts with H2SO4?
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    (Original post by alevelsguy)
    Where does it say that 2KOH reacts with H2SO4?
    It doesn't have to say it..

    The question is also testing your knowledge of the reactions of acids and bases.

    Sulphuric acid has two hydrogen ions that can react with the OH- ions of bases.

    So 1 mole of sulphuric acid reacts with 2 moles of KOH (which only provides one OH- ion per mole)
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    Ah fair dues, cheers!
 
 
 
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