Question

In 120 kg of mixed fertilizer, the amount of urea and potash are 60% and 40%
respectively .Let's complete, how many kgs of fertilizers of each type are present
in the mixed fertilizer​

Answer 1

Answer :

Kgs of urea in fertilizer = (60/100)*120

=72 Kg

Kgs of potash in fertilizer = 120 - 72

= 48 Kg.

Answer 2

Answer:

So now I guess that if you said you want N-P-K at 120-60-50 kg/ha rate that these rates are really for N-P-K and not N- P2O5 and K2O.

If not the conversion factors are:

P2O5 x 0.4364 = P

P x 2.20 = P2O5

K2O x 0.83 = K

K x 1.20 = K2O

These conversion factors are based on the atomic mass:

O = 15.99

K = 39.0983

P = 30.973

Ex.

K2O= = (2 x 39.0983) + 15.99 = 78.1966 + 15.99 = 94.1866

So there is 2 K in K2O

There is 78.1966 (K) in 94.1866 (K2O_X = 78.1966 / 94.1866 = 0.83

So the conversion factor is K2O x 0.83 = K

The other way around is 94.1866 / 78.1966 = 1.204 thus the conversion factor is

1.2 x K = K2O

You can do the same exercise for P2O5

Now there is two way to apply your fertilizers either by calculating the surface of 1 ha and the surface of your pots, or you calculate the weight of one ha and we know that you have 2 kg in each pot.

Now the weight of 1 ha of soil will greatly depend on the soil bulk density, the nature of soil (minreal or organic, its organic matter content ect…).

Since we don’t have the information about your soil we can consider that it is a typical mineral soil and its hectare –furrow slice of its surface soil will weight 2.2 million kilograms.

To simplify your calculation you can consider 2 million kilograms.

Now the determination of your need of N-P-K at 120-60-50 kg/ha rate I guess was performed on the basis