Practice Problems for Effective Calcium Carbonate Equivalent
Agronomy 354

The formula for calculating the Effective Calcium Carbonate Equivalent (ECCE) for agricultural limestone is:

ECCE = Calcium Carbonate Equivalent (CCE) x Fineness Factor.
Where:
CCE is determined in a laboratory (acid neutralizing factor)

Fineness Factor (Iowa Limestone Law) is calculated from:

_______  % material passing a 4-mesh screen             x 0.1         = _________

+   _______   % material passing an 8-mesh screen          x 0.3         = _________

+   _______   % material passing a 60-mesh screen          x 0.6         = _________

TOTAL Fineness Factor                      = _________

1. You have a liming material with a CCE of 80%. 100% of it passed a 4-mesh screen, 80% passed an 8-mesh screen, and 70% passed a 60-mesh screen. What is the fineness factor?
_100___  % material passing a 4-mesh screen             x 0.1         = ____10___

+   __80___   % material passing an 8-mesh screen          x 0.3         = ____24___

+   __70___   % material passing a 60-mesh screen          x 0.6         = ____42___

TOTAL Fineness Factor                      = ____76__

What is the % ECCE?

80% CCE x 0.76 Fineness Factor = 60.8%

How much of this material would you need to apply to supply 2500 lbs of ECCE/ac?

2500 lbs of ECCE/ac / 0.608 = 4125 lbs/ac liming material
2. How much CaCO3 would you apply to a kilogram of soil in a flower pot to neutralize 2 meq acidity/100 g of soil?
(Atomic wts: Ca = 40, C = 12, O = 16, H = 1)
meq wt of CaCO3 = molecular wt / 2 (valence of the cation Ca is ++)

100 / 2 = 50 mg/meq

2 meq acidity will require 2 meq of lime

Therefore, 2 meq of lime (pure CaCO3) = 100 mg CaCO3 required per 100 g soil,

or 1000 mg CaCO3 required per 1000 g (kg) of soil  (this is 1 g CaCO3 per kg of soil)

3. How much CaO would you apply to the soil in question 2?
meq wt of CaO = molecular wt / 2 (valence of the cation Ca is ++)

56 / 2 = 28 mg/meq

Therefore, 2 meq of CaO = 56 mg CaO required per 100 g soil

or 560 mg CaO required per 1000 g (kg) of soil (this is a little more than 1/2 g CaO per kg of soil)

4. How much Ca(OH)2 would you apply to an acre-6" of the soil in question 2?
meq  wt of Ca(OH)2 = molecular wt / 2 (valence of cation Ca is ++)

74 / 2 = 37 mg/meq

Therefore, 2 meq of Ca(OH)2 = 74 mg Ca(OH)2 required per 100 g soil

Conversion of mg per 100 g x 20 = lbs/2,000,000 lbs soil (acre furrow slice) (NOTE: As discussed in lecture, this factor converts mg per 100 g soil to lbs per acre.)

74 x 20 = 1480 lbs of Ca(OH)2 required per acre

5.  What is the % CCE of Ca(OH)2?
CCE compares the relative effectiveness of the liming sample vs. pure CaCO3.
It takes 50 mg of CaCO3 to neutralize 1 meq of acidity but only 37 mg of Ca(OH)2.  Therefore, Ca(OH)2 has a neutralizing valve greater than 100%.

50 / 37 x 100 (to get %) = 135%

6. Your soil test report calls for 1900 lbs of ECCE per acre for a 6-inch depth but you wish to incorporate the lime to an 8-inch depth. How much lime is required?
Lime is adjusted proportionally with depth and most recommendations are for a 6-inch depth. Therefore, if the lime is mixed into more soil, more lime is required. The same procedures are used if the lime is to be incorporated into shallower depths where less limed is needed. One can easily set up a proportion and solve for X. Thus, 6 is to 1900 as 8 is to X. Cross-multiply and solve for X:

6/8 = 1900/X; X = 2533 lbs of ECCE

7. Your local quarry has a liming material that is 87% ECCE and costs \$13.00/ton ECCE delivered and spread on your field. Your soil test report calls for 3400 lbs/ac of ECCE. a) If you are applying lime to 20 acres, how much should you pay for the lime? b) How many tons of limestone must the quarry deliver to your field to meet this requirement?

a) As is often the case, the liming material from the quarry is not 100% effective. In this instance where the material is only 87% effective, more that 3400 lbs/ac must be applied.

3400/0.87 = 3908 lbs of this liming material is required to equal 3400 lbs of ECCE (but Iowa Limestone Law requires that the quarry price their material based on ECCE). Thus the cost of lime per acre is 3400/2000 = 1.7 tons/ac X \$13.00/ton = \$22.10/acre

This is applied to a 20-ac field.

20 ac X \$22.10/ac = \$442.00

b) To get the correct amount of lime applied, 3908 lbs must be applied per acre (which gives 3200 lbs ECCE becasue it is only 87% effective).

3908 lbs/ac X 20 ac = 78160 lbs limestone from this quarry

In tons:

78160 lbs/2000 = 39.08 tons

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