Firstly since 20% of Na3
mixture consists of contaminants, only 8g of Na3
is available. 164 g of it gives 31g P and thus 8g of it should give 1.512g P
Now 132 g of (NH4
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
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
, 10g of the Na3
mixture given and 7.83g of KNO3
Thus the net weight of fertilizer is 28.52g