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### Topic: Problem of the week - 25/03/2013  (Read 23307 times)

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#### Borek

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##### Problem of the week - 25/03/2013
« on: March 25, 2013, 12:19:46 PM »
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 26, 2013, 05:17:10 AM »
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
1.512g
5.6g
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.

#### DrCMS

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##### Re: Problem of the week - 25/03/2013
« Reply #2 on: March 26, 2013, 11:49:08 AM »
I got a value of 29.0g

#### Sunil Simha

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##### Re: Problem of the week - 25/03/2013
« Reply #3 on: March 26, 2013, 12:19:06 PM »
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

#### Rutherford

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##### Re: Problem of the week - 25/03/2013
« Reply #4 on: March 29, 2013, 02:00:54 PM »
I got 22.3g.

#### curiouscat

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

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?

#### Sunil Simha

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##### Re: Problem of the week - 25/03/2013
« Reply #6 on: April 02, 2013, 01:57:40 PM »

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?)

#### DrCMS

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##### Re: Problem of the week - 25/03/2013
« Reply #7 on: April 02, 2013, 06:31:09 PM »
After adjusting my sums to include 2 nitrogens from the ammonium sulphate I get 23.4g.

#### Borek

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

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##### Re: Problem of the week - 25/03/2013
« Reply #9 on: April 08, 2013, 10:08:23 AM »
I didn't expect any problems with this question, but apparently there are no problems easy enough to not allow for a mistake

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.

#### AWK

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