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    0.5g of impure ammonium chloride is warmed with an excess of sodium hydroxide solution. The ammonia liberated is absorbed in 25cm3 of 0.2mol dm-3 sulphuric acid. The excess of sulphuric acid requires 5.64cm3 of 0.2 moldm-3 sodium hydroxide solution for titration. Calculate the percentage of ammonium chloride in the original sample.

    Here are the calculations I have done:

    NH4Cl + NaOH + NH3 + H2O + NaCl


    2NH3 + H2SO4 + (NH4)2SO4

    Moles of sulphuric acid = (25x0.2)/1000 = 0.005moles


    H2SO4 + NaOH + H2O + NaHSO4

    Number of moles of sodium hydroxide = (5.64x0.2)/1000 = 0.001128 moles


    Amount of sulphuric acid reacted = 0.005-0.001128=0.003872

    What should I do next? I do not understand what it means when it says ammonia liberated is absorbed into sulphuric acid.

    Please help!

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    So now you know the number of moles of sulphuric acid reacted, you know that twice this number of moles gives you the number of moles of NH3 produced. Number of moles of NH3 = 2*0.003872 = 0.007744

    Since 1 mole NH4Cl -> 1 mole NH3

    Therefore the number of moles of NH4Cl in the samle is 0.007744

    This weighs 0.007744*(14+4*1+35.5) = 0.41304g

    Therefore percentage by mass = (0.41304/5) *100 = 83% to 2sf
 
 
 
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