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    ok, this is from an Edexcel past paper (June 2002)

    A fertiliser is known to contain ammonium sulphate, (NH4)2SO4, as the only ammonium salt

    A sample weighing 3.80g was dissolved in water and the volume made to 250cm^3. To 25cm^3 portions of this solution about 5cm^3 (an excess) of aq methanal was added. The following reaction took place:

    4(NH4+) + 6HCHO = C6H12N4 + 4H+ + 6H2O

    The liberated acid was titrated directly with 0.100 mol dm^-3 aq sodium hydroxide. The average volume required was 28.0cm^3. Calculate the percentage of ammonium sulphate in the fertiliser


    The answer is 48.6% - but I really dont understand the mark scheme
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    (Original post by Lupin Loopy)
    ok, this is from an Edexcel past paper (June 2002)

    A fertiliser is known to contain ammonium sulphate, (NH4)2SO4, as the only ammonium salt

    A sample weighing 3.80g was dissolved in water and the volume made to 250cm^3. To 25cm^3 portions of this solution about 5cm^3 (an excess) of aq methanal was added. The following reaction took place:

    4(NH4+) + 6HCHO = C6H12N4 + 4H+ + 6H2O

    The liberated acid was titrated directly with 0.100 mol dm^-3 aq sodium hydroxide. The average volume required was 28.0cm^3. Calculate the percentage of ammonium sulphate in the fertiliser


    The answer is 48.6% - but I really dont understand the mark scheme
    I find the best thing for these questions is to work backwards.

    Sodium hydroxide donates one (OH-) ion during reactions. This means 4 moles would react with the 4 (H+) ions. 28 cm of 0.1 mol NaOH contains 0.028 moles of reactant so there must be the same moles of (H+)

    This in turn reacts with 4 NH4+ ions, as in this reaction each ammonium fertiliser molecule donates two ions to the reaction, it has a 2:4 ratio, hence 0.05 molecules of fertiliser reacts with 1 OH- ion, this means there must be 0.0014 moles of fertiliser in each sample.

    If it was pure fertiliser then in each 25 cm^3 sample there would be (.38/132) moles, this equals .00287878787 moles, however we know there is only .0014 mole in the sample

    So...
    Working out percentage by mass
    (0.0014/0.00287878787 *100) = 48.6

    Hope that helps!
 
 
 
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