132 FERTIUTY AND rERTILIZEK HINTS 



The nitrate of soda is given as carrying an equivalent of 18.84 

 per cent, of ammonia. To convert ammonia into nitrogen we 

 must multiply by the factor 0.823. 



18,84 Jo ammonia X 0.823 = 15.5 % nitrogen. 



Then : 



2 X 15 = 30 o lbs. available phosphoric acid 



2 X 2.5 = 5.0 lbs. nitrogen from dissolved bone superphosphates 



1 X 15-5 = 15-5 lbs. nitrogen from nitrate of soda 



0.5 X 64 = 32.0 lbs. potash. 



The percentages in this mixture would be : 

 Available phosphoric acid lbs. 30 -^ 350 = 8.57 Jr available phosphoric acid 



Nitrogen lbs. 20.5 -j- 350 = 5.85 Jo nitrogen 



Potash lbs. 32 H- 350 = 9. 14 Jc potash. 



How to Calculate Amounts from Known Percentages. — If 2.000 

 pounds of a mixture analyzing 



Available phosphoric acid 7 per cent. 



Nitrogen 5 per cent. , and 



Potash 6 per cent. 



is desired from 



Acid phosphate analyzing 16 Jo available phosphoric acid. 

 Calcium cyanamid " 17 Jo nitrogen, and 



Muriate of potash " 50 ^ir potash. 



it may be calculated in the following way : 



First find out the number "of pounds of available phosphoric 

 acid, nitrogen and potash that would be required. Since 2,000 

 IS 20 times 100 we may multiply the percentages by 20. 

 20 X 7 ( ^ avail, phos. acid) = 140 lbs. avail, phos. acid required for 2,000 lbs. 

 20 X 5 (% nitrogen ) = loolbs. nitrogen " " " " 



20X6 (% potash) — 1 20 lbs. potash 



To determine the number of pounds of acid phosphate, calcium 

 cyanamid and muriate of potash needed to give the analysis 

 desired, we may divide the pounds of available phosphoric 

 acid, nitrogen and potash by the percentages that the materials 

 analyzed. 

 Avail, phos. acid lbs. 140 -r- 16 V = S75 lbs. acid phosphate required 

 Nitrogen " lot^ -i- 17 ^ := 588 lbs. calcium cynamid required 



Potash ■■ 120-j- 50 Jc ^= 240 lbs. muriate of potash required 



Total = 1,703 lbs 



