You decide that you want an ...

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.