[Borek]

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?

[Sunil Simha]

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  1.512g
N  5.6g
K  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]

I got a value of 29.0g

[Sunil Simha]

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]

I got 22.3g.

[curiouscat]

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]

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]

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

[Borek]

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 

[DrCMS]

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]

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