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2002 - Written Paper - Q10(d), Q14 (ii)

Sammie

Posted 19 April 2006 - 03:12 PM

okies I was doign the 2002 past paper and got stuck on these!

10d.

Under certan conditions 200kg of hydrogen reacts with excess nirogen in the Haber Process to produce 650kg of Ammonia

Calculate the percentage yield of ammonia

and

14ii

1.02 x 10 exp -5 mol of iodine reacted with 3.0cmcubed of sodium thiosulphate soluition.

Use this information to calculate the conventration of the sodium tiosulphate in mol l-1

ad absurdum

Posted 19 April 2006 - 09:20 PM

QUOTE(Sammie @ Apr 19 2006, 04:12 PM) View Post

okies I was doign the 2002 past paper and got stuck on these!

10d.

Under certan conditions 200kg of hydrogen reacts with excess nirogen in the Haber Process to produce 650kg of Ammonia

Calculate the percentage yield of ammonia
Well the reaction between hydrogen and nitrogen in the Haber Process is:
3H2 + 2N2 --> 2NH3

From this we can see that 3 mole of hydrogen reacts with 2 mole of nitrogen to give 2 mole of ammonia. In this question, the nitrogen is in excess, so the hydrogen is going to decide how much ammonia is formed.
One mole of hydrogen is 2g. We have 200kg = 200000g; which is 100000 mole of hydrogen
If 3 mole of hydrogen gives 2 mole of ammonia
100000 mole of hydrogen gives (2/3 * 100000) mole of ammonia

So we have 200000/3 mole of ammonia as the theoretical yiled
The actual yiled is 650kg

One mole of ammonia weighs 20g, 650kg is 32500 mole
We should get 200000/3 mole, we really get 32500 mole
The percentage yeild is (actual yield/theoretical yield)*100 = [32500/(200000/3)] * 100 = 48.75%


The next one, well from the equation given in the question, one mole of iodine reacts with two mole of thiosulphate. This means that 1.2 * 10^-5 mole of iodine will react with 2.4 * 10^-5 mole of thiosulphate ions.
Using n = c * V (where n is the number of moles, c is the concentration in moles per litre and V is the volume in litres, so we'll need to change 3cm^3 into 0.003 litres)
rearrange to get c = n / V = (2.4 * 10^-5) / 0.003 = 0.008 mol l^-1

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