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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

You need to prepare a 1.5:1:2 (mass ratio) NPK plant fertilizer, using 26.4 g of (NH_{4})_{2}SO_{4}, 10g of Na_{3}PO_{4} and 20 g of KNO_{3}. How much of the fertilizer can you prepare, if Na_{3}PO_{4} contains 20% of inert contaminants?

Firstly since 20% of Na_{3}PO_{4} mixture consists of contaminants, only 8g of Na_{3}PO_{4} is available. 164 g of it gives 31g P and thus 8g of it should give 1.512g P
Now 132 g of (NH_{4})_{2}SO_{4} gives 28g N and hence 26.4 g of it should give 5.6g N.
Finally 101g of KNO_{3} 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 (NH_{4})_{2}SO_{4}, 10g of the Na_{3}PO_{4} mixture given and 7.83g of KNO_{3}.
Thus the net weight of fertilizer is 28.52g.

I got a value of 29.0g

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

I got 22.3g.

Finally 101g of KNO_{3} 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?

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?)

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

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

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.

http://en.wikipedia.org/wiki/NPK_rating