Question

A water treatment plant treats 25 MLD water with a natural alkalinity of 4.0 mg/L (as CaCO$_3$). It is estimated that, during coagulation of this water, 450 kg/day of calcium bicarbonate (Ca(HCO$_3$)$_2$) is required based on the alum dosage. Consider the atomic weights as: Ca=40, H=1, C=12, O=16. The quantity of pure quick lime, CaO (in kg) required for this process per day is ______ (rounded off to 2 decimal places).

Hint:

Understanding the stoichiometric relationships in chemical reactions is crucial for calculating the required amounts of reactants or products in industrial processes such as water treatment.

Answer 1

Step 1: Calculate the molar mass of calcium bicarbonate and quick lime. - Molar mass of Ca(HCO$_3$)$_2$: \[ \text{Ca(HCO}_3)_2 = 40 + 2 \times (1 + 12 + 3 \times 16) = 40 + 2 \times 61 = 162 \text{ g/mol} \] - Molar mass of CaO: \[ \text{CaO} = 40 + 16 = 56 \text{ g/mol} \] Step 2: Determine the stoichiometry of the reaction converting Ca(HCO$_3$)$_2$ to CaO. \[ \text{Ca(HCO}_3)_2 \rightarrow \text{CaO} + \text{CO}_2 + \text{H}_2\text{O} \] From the stoichiometry, 1 mole of Ca(HCO$_3$)$_2$ produces 1 mole of CaO. Step 3: Convert the daily usage of Ca(HCO$_3$)$_2$ to moles, then to CaO. \[ \text{Daily usage in moles of Ca(HCO}_3)_2 = \frac{450,000 \text{ g/day}}{162 \text{ g/mol}} \approx 2777.78 \text{ moles/day} \] \[ \text{Daily requirement of CaO in moles} = 2777.78 \text{ moles/day} (since the ratio is 1:1) \] \[ \text{Daily requirement of CaO in kg} = 2777.78 \text{ moles/day} \times 56 \text{ g/mol} / 1000 = 155.56 \text{ kg/day} \]