I just have a question : In case I have NPK 20:20:20 and I want to use it in the foliar spray application by 200 ppm how much NPK I have to take and dissolve it in 1 Liter to make a stock solution. Regards
Hi Sir! Thank you for that question. If you have a fertilizer grade of 20-20-20 NPK, it means that you have 200g of pure NPK in 1 kg. So to calculate the amount of fertilizers to use for 200pm spray application, you may use this formula: Amount of fertilizer = (desired ppm*amount of water)/ product concentration Amount of fertilizer (20-20-20) = (200*1L)/200g/L (as 1kg = 1L, so instead to use kg, convert it to L) Amount of fertilizer (20-20-20) = 200L/200 g/L Amount of fertilizer (20-20-20) = 1g (which means when you add 1 g of fertilizer (20-20-20) to 1L of water, the resulting solution will be 200pm.) I hope this helps you. God bless!
@@rheajoyflora2054 I highly appreciate your fast reply, but what will be the method of calculation when P represented by P2O2 and K represented by K2O which is common, I mean is not more accurate to find out the exact ( the net amount) percentage of phosphorous and potassium within a fertilizer bag. As in my case, the real percentage of 20-20-20 gone be like 20-11.6-8.8 as I calculated below: For Nitrogen 20%= 20% net N For Phosphorous: 20% = 8.8% net P For Potassium: 20%=11.6 net K P2O5= (2*39.098) +(5*15.999 = 78.176+15.999) = 61.948+79.995=141.9434 = (61.948/141.9434)*100 =44% P = 0.44 P =20%*0.44= 8.8 Net P= 8.8% K2O= (2*39.098) +15.999 = 78.176+15.999=94.195 =78.176/94.195 =83% K =0.83 K =20%*0.83=11.6% Net K= 11.6% The question here is with a bag containing 20-11.6-8.8 how can I prepare 200 ppm of NPK for foliar spray?
@@hassann.muhamed929 Based on the net amount of P and K in the fertilizer, the bag of fertilizer contain 20% N, 8.8% P, and 16.6% K (not 11.6% K you might miscalculated this one). As I mentioned in my previous reply, you will need 1g of fertilizer (20-20-20) per liter of water as your stock solution to achieve 200 ppm of N (it is only Nitrogen that you will get an exact 200 ppm of N because the computation was based on N having the highest concentration of element). However, when you calculated the net amount of P and K in the fertilizer, the elemental P and K in the fertilizer is 8.8% and 16.6%, respectively. Now, when 1g fertilizer (20-20-20, with net P of 8.8% and K of 16.6%) is diluted to 1 L of water, it will result to 200 ppm N. However, the amount of fertilizer will only supply 88 ppm of P, and 166 ppm of K in the solution. Since you need 200 ppm for all elements, you will be needing single element fertilizers to make up P (112 ppm), and K (34 ppm) which is not supplied by 1g fertilizer (20-20-20) to be diluted in 1 liter. I am not familiar as to what type of fertilizers (single element) you have in your country so, I could not calculate the amount for you. By the way, if you need more volume of stock solution, say you want it to be diluted in 16 liters of water (as in 16-liter capacity sprayer), then you need 16 g (fertilizer 20-20-20) plus other fertilizers (single element) to reach the desired 200 ppm NPK. I hope this helps you.
@@rheajoyflora2054 you are right, its possible only with single element fertilizers, with a balanced fertilizer as in my case 20:20:20 most likely the NPK production is based on the pure elements (p and K not P2O2 and K2O) as the company informed me about their product. Thanks again
Determine mo ang recommended fertilizer rate per hectare, and compute for the amount of fertilizer type you need based on the recommended fertilizer rate. Then, determine the number of plants/hills per hectare nang target crop to be planted in pots. Then divide the amount of each fertilizer needed per hectare by the number of plants per hectare then you will get the amount of fertilizers to be applied per plant/ hill. Then that amount will be used per pot. If you still didn't get my explanation, I will show you the formula.
Already done of watching videos last week ma'am, THANK YOU!!😊
I just have a question :
In case I have NPK 20:20:20 and I want to use it in the foliar spray application by 200 ppm how much NPK I have to take and dissolve it in 1 Liter to make a stock solution.
Regards
Hi Sir! Thank you for that question. If you have a fertilizer grade of 20-20-20 NPK, it means that you have 200g of pure NPK in 1 kg. So to calculate the amount of fertilizers to use for 200pm spray application, you may use this formula:
Amount of fertilizer = (desired ppm*amount of water)/ product concentration
Amount of fertilizer (20-20-20) = (200*1L)/200g/L (as 1kg = 1L, so instead to use kg, convert it to L)
Amount of fertilizer (20-20-20) = 200L/200 g/L
Amount of fertilizer (20-20-20) = 1g (which means when you add 1 g of fertilizer (20-20-20) to 1L of water, the resulting solution will be 200pm.)
I hope this helps you. God bless!
@@rheajoyflora2054 I highly appreciate your fast reply, but what will be the method of calculation when P represented by P2O2 and K represented by K2O which is common, I mean is not more accurate to find out the exact ( the net amount) percentage of phosphorous and potassium within a fertilizer bag.
As in my case, the real percentage of 20-20-20 gone be like
20-11.6-8.8 as I calculated below:
For Nitrogen 20%= 20% net N
For Phosphorous: 20% = 8.8% net P
For Potassium: 20%=11.6 net K
P2O5= (2*39.098) +(5*15.999 = 78.176+15.999) = 61.948+79.995=141.9434
= (61.948/141.9434)*100 =44% P = 0.44 P
=20%*0.44= 8.8
Net P= 8.8%
K2O= (2*39.098) +15.999 = 78.176+15.999=94.195
=78.176/94.195
=83% K =0.83 K
=20%*0.83=11.6%
Net K= 11.6%
The question here is with a bag containing 20-11.6-8.8 how can I prepare 200 ppm of NPK for foliar spray?
@@hassann.muhamed929 Based on the net amount of P and K in the fertilizer, the bag of fertilizer contain 20% N, 8.8% P, and 16.6% K (not 11.6% K you might miscalculated this one). As I mentioned in my previous reply, you will need 1g of fertilizer (20-20-20) per liter of water as your stock solution to achieve 200 ppm of N (it is only Nitrogen that you will get an exact 200 ppm of N because the computation was based on N having the highest concentration of element). However, when you calculated the net amount of P and K in the fertilizer, the elemental P and K in the fertilizer is 8.8% and 16.6%, respectively. Now, when 1g fertilizer (20-20-20, with net P of 8.8% and K of 16.6%) is diluted to 1 L of water, it will result to 200 ppm N. However, the amount of fertilizer will only supply 88 ppm of P, and 166 ppm of K in the solution. Since you need 200 ppm for all elements, you will be needing single element fertilizers to make up P (112 ppm), and K (34 ppm) which is not supplied by 1g fertilizer (20-20-20) to be diluted in 1 liter. I am not familiar as to what type of fertilizers (single element) you have in your country so, I could not calculate the amount for you.
By the way, if you need more volume of stock solution, say you want it to be diluted in 16 liters of water (as in 16-liter capacity sprayer), then you need 16 g (fertilizer 20-20-20) plus other fertilizers (single element) to reach the desired 200 ppm NPK.
I hope this helps you.
@@rheajoyflora2054 you are right, its possible only with single element fertilizers, with a balanced fertilizer as in my case 20:20:20 most likely the NPK production is based on the pure elements (p and K not P2O2 and K2O) as the company informed me about their product. Thanks again
Already watching this video last week ma'am 😊
done watching , thank you ma'am ❣️
Done ma'am. Thank you.its a big help gd sa problem solving namun in fertilizer computation.
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Done watching ma'am 😘😇. Thank you so much its a big help for us ma'am
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Already watched this lastweek maam But I was unable to comment!apology and thankyou!
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Mam pano po kapag pot experiment po. Ano po ang right way to convert into grams per pot?
Determine mo ang recommended fertilizer rate per hectare, and compute for the amount of fertilizer type you need based on the recommended fertilizer rate. Then, determine the number of plants/hills per hectare nang target crop to be planted in pots. Then divide the amount of each fertilizer needed per hectare by the number of plants per hectare then you will get the amount of fertilizers to be applied per plant/ hill. Then that amount will be used per pot. If you still didn't get my explanation, I will show you the formula.
@@rheajoyflora2054 Thank you very much po 😊
@@rheajoyflora2054 Mam, medyo confused parin po, sorry po. Pwede po makita ang pinaka formula po?
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Good Evening Maam. Pwedi po maka request ng handouts about your topic? Thanks and God bless po.
Sorry for the very late reply. I just noticed your message. I could have sent you if I've this.
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Ma'am Akon ni sa ma'am! Nalipat ko mag switch acc♥️
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