Chemical equation and calculation for remineralization system

Question

Dear all,
I am designing a remineralization system(Lime and CO2 gas injection to BWRO permeate ). So, our company received consultant from a vendor.
Below chemical equation and calculation had calculated from a  vendor.
However, we don't know how to calculate below data "As such required: 39.22, 46.64, 85.86", "Actual alkalinity produced: 47.567, 87.5675".
If you know well chemical equations and this calculation, please let me know how to calculate.
Thank you.   
 
Dose Calculation for Lime + CO2
                                                    Ca(OH)2 + 2CO2 = Ca(HCO3)2
                                                          74           88            162
As such required:                         39.22      46.64        85.86
As CaCO3:                                        53           53              53
Dosages as Such provided:           40           50          92.045
Actual alkalinity produced:                        47.567      87.5675

J.J.P. (Hans) Zijlstra · Accepted Answer

You decide that you want an alkalinity of 53 mg/L of calcium carbonate.
You add the molar masses of Ca = 40 g/mole, C = 12 g/mole and three times O = 16 g/mole to find that 1 mole of CaCO3 or calcium carbonate weighs 100 g.
You decide that you need 0.53 mole of calcium carbonate in 1000 L.
With 1 mole of Ca(OH)2 and 2 moles of CO2 you can make 1 mole of Ca(HCO3)2, or 1 mole of CaCO3 alkalinity.
While also knowing that 1 mole of H weighs 1 g, you now calculate the weights of 1 mole Ca(OH)2, 2 moles of CO2 and 1 mole of Ca(HCO3)2, which are 74, 88, and 162 g, respectively.
You only need 0.53 mole, or 0.53*74 = 39.22 g of Ca(OH)2, 0.53*88 = 46.46 g of CO2, to make 0.53*162 = 85.86 g of Ca(HCO3)2, equivalent to the alkalinity of 53 g of CaCO3.
So you weigh, as a matter of convenience, 40 g instead of 39.22 g of Ca(OH)2 and you add 50 g of CO2.
If 50 g of CO2 react instead of 46.46 g, then you produce (50/46.46)*85.86 = 92.045 g of Ca(HCO3)2, but then you have not enough Ca(OH)2.
So, 40 g of Ca(OH)2 shall react, then you produce (40/39.22)*85.86 = 87.568 g of Ca(HCO3)2 and some CO2 is left.
In attached excel file you find the calculations also.

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