JohnDeere8530
Member
I have a 800t pile of organic plant material from a nursery on my farm. An analysis report states in it has 2500mmol/kg DS of Nitrogen in it.
I wish to apply this next year on the freshly drilled OSR crop (have EA permission) However I am struggling with the amount of Nitrogen I will be applying
2500mmol/kg… I take that as to mean 2500 x10^-3 moles of Nitrogen per 1kg of dry material
Therefore this equates to 2.5mol/kg DM
Nitrogen has a RAM of 14 so that equates to 35g of N/kg (From memory of my chemistry days)
Therefore applying 800t (800,000kg)
=800,000*35grams of N
=28,000,000gm of N
=28t of N
But that assumes it is 100% dry, ( I think by looking at lab report)
So if it is 10% dry that is
approximately 2.8t of N in the 800t pile.
Is that correct?
It is possible I may be a power of 10 out somewhere though !!


I wish to apply this next year on the freshly drilled OSR crop (have EA permission) However I am struggling with the amount of Nitrogen I will be applying
2500mmol/kg… I take that as to mean 2500 x10^-3 moles of Nitrogen per 1kg of dry material
Therefore this equates to 2.5mol/kg DM
Nitrogen has a RAM of 14 so that equates to 35g of N/kg (From memory of my chemistry days)
Therefore applying 800t (800,000kg)
=800,000*35grams of N
=28,000,000gm of N
=28t of N
But that assumes it is 100% dry, ( I think by looking at lab report)
So if it is 10% dry that is
approximately 2.8t of N in the 800t pile.
Is that correct?
It is possible I may be a power of 10 out somewhere though !!
