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A2 Chemistry Really Hard Q (edexcel) watch

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    This is from a chem synoptic paper (jun 02 - Q1a)

    1) A fertiliser is known to contain ammonium sulphate, (NH4)2SO4, as the only ammonium salt. This Q concerns methods for the determination of the ammonium and sulphate ions.

    A sample weighing 3.80g was dissolved in water and the volume made to 250cm3. To 25 cm3 portions of this solution about 5cm3 (excess) of methanal(aq) was added. The following reaction took place:

    4NH4*(aq) + 6HCHO(aq) --> C6H12N4(aq) + 4H*(aq) + 6H20

    the liberated acid was titrated directly with 0.100 mol dm-3 aq NaOH. The average vol required was 28.0cm3.

    a) Calculate the %age of ammonium sulphate in the fertiliser.

    OK, i know to work out the no. moles of NaOH =28 x 0.100 / 1000

    = 2.8 x 10 ^-3

    Then i'm stuck. But i know the answer is 48.6% if that helps
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    Here is a very similar question and hope it helps:

    Ammonium salts are widely used as fertilisers. One standard method for the analysis of ammonium salts (except the chloride) is to react them in a solution of methanal, HCHO. This forms a netural organic compound together with an acid which can be titrated with the standard alkali. For ammonium nitrate the equation for the reaction is:

    4 NH4NO3 + 6HCHO => (CH2)6N4 + 4 HNO3 + 6H20

    15.0 g of a fertiliser containing ammoniun nitrate as the only ammonium salt was dissolved in water and the solution made up to 1.00 dm3 with pure water,

    25.0 cm3 portions of this solution were then treated with saturated aqueous methanol solution and allowed to stand for a few minutes.

    The liberated nitric acid was then titrated with 0.100 mol dm-3 NaOH solution. The volume of NaOH solution used was 22.3 cm3. What percentage by mass of the fertiliser was ammonuium nitrate?

    Equation 2 is (i think) : HN03 + NaOH => NaNO3 + H20

    So moles of NaOH = (0.100 * 22.3) / 1000 = 0.00223

    Moles of HNO3 = 0.00223 (same)

    Moles in 1.00dm3 (1000 cm3) = 0.00223 * 40 = 0.0292

    Mass = 0.0292 * 80 = 7.136

    % = 7.136 / 15 * 100 = 47.573 (book gives 47.6%)
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    (Original post by i_love_maths)
    This is from a chem synoptic paper (jun 02 - Q1a)

    1) A fertiliser is known to contain ammonium sulphate, (NH4)2SO4, as the only ammonium salt. This Q concerns methods for the determination of the ammonium and sulphate ions.

    A sample weighing 3.80g was dissolved in water and the volume made to 250cm3. To 25 cm3 portions of this solution about 5cm3 (excess) of methanal(aq) was added. The following reaction took place:

    4NH4*(aq) + 6HCHO(aq) --> C6H12N4(aq) + 4H*(aq) + 6H20

    the liberated acid was titrated directly with 0.100 mol dm-3 aq NaOH. The average vol required was 28.0cm3.

    a) Calculate the %age of ammonium sulphate in the fertiliser.

    OK, i know to work out the no. moles of NaOH =28 x 0.100 / 1000

    = 2.8 x 10 ^-3

    Then i'm stuck. But i know the answer is 48.6% if that helps
    hey i done this exact question before hope this helps



    1)first u calculate #moles of NaOH which is 2.8 x10 -3

    2)then calculate #moles of H+ = 2.8 x 10 -3 = 1.4 x 10 -3 (in 25cm3)

    3)fine the # moles of H+ in 250cm3 = 1.4 x10 -3 x 10 = 1.4x 10 -2

    4) then u find the mass of the ammonium sulphate
    we know #moles = 1.4x 10 -2
    also that the RMM = 132
    so mass = #moles x RMM = (1.4 x 10 -2) x 132
    mass = 1.848g

    5) % of ammonium sulphate in fertiliser = 1.848/ 3.8 X 100% = 48.6%

    hope that helps ya!!
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    Thanx!!!

    I guess it wasn't as hard as i was tryin to make it!!!!!
 
 
 
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