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Author Topic: Problem of the week - 25/03/2013  (Read 9578 times)

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Borek

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Problem of the week - 25/03/2013
« on: March 25, 2013, 06:19:46 AM »

You need to prepare a 1.5:1:2 (mass ratio) NPK plant fertilizer, using 26.4 g of (NH4)2SO4, 10g of Na3PO4 and 20 g of KNO3. How much of the fertilizer can you prepare, if Na3PO4 contains 20% of inert contaminants?
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Sunil Simha

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Re: Problem of the week - 25/03/2013
« Reply #1 on: March 25, 2013, 11:17:10 PM »

Firstly since 20% of Na3PO4 mixture consists of contaminants, only 8g of Na3PO4 is available. 164 g of it gives 31g P and thus 8g of it should give 1.512g P

Now 132 g of (NH4)2SO4 gives 28g N and hence 26.4 g of it should give 5.6g N.

Finally 101g of KNO3 give 39 g K and hence 20g of it must give 7.72g of K.

So now to summarize what is available to us
:rarrow: 1.512g
:rarrow: 5.6g
:rarrow: 7.72g

Thus by trial and error, we can observe that P is the limiting factor. Thus 1.512g P needs 1.5 times N and 2 time the K i.e. 2.268g N and 3.024g K.

These can be obtained from 10.69g of (NH4)2SO4, 10g of the Na3PO4 mixture given and 7.83g of KNO3.

Thus the net weight of fertilizer is 28.52g.
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DrCMS

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Re: Problem of the week - 25/03/2013
« Reply #2 on: March 26, 2013, 05:49:08 AM »

I got a value of 29.0g
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Sunil Simha

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Re: Problem of the week - 25/03/2013
« Reply #3 on: March 26, 2013, 06:19:06 AM »

I got a value of 29.0g

I think our answers differ because I have rounded off the atomic masses to whole numbers and have, at some places, not really taken the significant digits to consideration. But they are close enough. :) But if there are any mistakes in my calculations, then please tell me so that I can correct them.

Thanks
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Rutherford

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Re: Problem of the week - 25/03/2013
« Reply #4 on: March 29, 2013, 08:00:54 AM »

I got 22.3g.
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curiouscat

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Re: Problem of the week - 25/03/2013
« Reply #5 on: April 01, 2013, 09:33:32 AM »


Finally 101g of KNO3 give 39 g K and hence 20g of it must give 7.72g of K.

Aren't you forgetting that KNO3 provides N in addition to K?
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Sunil Simha

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Re: Problem of the week - 25/03/2013
« Reply #6 on: April 02, 2013, 07:57:40 AM »


Aren't you forgetting that KNO3 provides N in addition to K?

I'm really sorry. :-[
Thanks a lot sir.
That makes 8.37g N available to us. So the final answer should be around 23g of fertilizer.(right?)
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DrCMS

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Re: Problem of the week - 25/03/2013
« Reply #7 on: April 02, 2013, 12:31:09 PM »

After adjusting my sums to include 2 nitrogens from the ammonium sulphate I get 23.4g.
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Borek

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Re: Problem of the week - 25/03/2013
« Reply #8 on: April 08, 2013, 02:09:47 AM »

Sorry to keep you waiting, I was away for Easter and forgot to take the charger - so my access to the web last Monday was limited. Then I was quite busy since getting back home.

23.4 g is the correct answer.

I didn't expect any problems with this question, but apparently there are no problems easy enough to not allow for a mistake  ;D
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DrCMS

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Re: Problem of the week - 25/03/2013
« Reply #9 on: April 08, 2013, 04:08:23 AM »

I didn't expect any problems with this question, but apparently there are no problems easy enough to not allow for a mistake  ;D

Yes indeed.  How I managed to use the correct formula to calculate the molar mass of ammonium sulphate but then only took one N from it I'm not sure but I did.  Doing that I got a value quite close to Sunil Simha who'd made a different mistake and we reinforced each others wrong answer rather than seeing our original errors.  Two wrong do not make a right but can quite often can make a right mess.
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AWK

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Re: Problem of the week - 25/03/2013
« Reply #10 on: April 08, 2013, 09:00:07 PM »

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