# Chemical Forums

## Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on March 25, 2013, 12:19:46 PM

Title: Problem of the week - 25/03/2013
Post by: Borek 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?
Title: Re: Problem of the week - 25/03/2013
Post by: Sunil Simha 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
P  :rarrow: 1.512g
N  :rarrow: 5.6g
K  :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.
Title: Re: Problem of the week - 25/03/2013
Post by: DrCMS on March 26, 2013, 11:49:08 AM
I got a value of 29.0g
Title: Re: Problem of the week - 25/03/2013
Post by: Sunil Simha 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
Title: Re: Problem of the week - 25/03/2013
Post by: Rutherford on March 29, 2013, 02:00:54 PM
I got 22.3g.
Title: Re: Problem of the week - 25/03/2013
Post by: curiouscat 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?
Title: Re: Problem of the week - 25/03/2013
Post by: Sunil Simha 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?)
Title: Re: Problem of the week - 25/03/2013
Post by: DrCMS on April 02, 2013, 06:31:09 PM
After adjusting my sums to include 2 nitrogens from the ammonium sulphate I get 23.4g.
Title: Re: Problem of the week - 25/03/2013
Post by: Borek 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  ;D
Title: Re: Problem of the week - 25/03/2013
Post by: DrCMS 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  ;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.
Title: Re: Problem of the week - 25/03/2013
Post by: AWK on April 09, 2013, 03:00:07 AM
http://en.wikipedia.org/wiki/NPK_rating