To find the quantity of quick lime required, we need to calculate the amount of lime required to neutralize the alkalinity in the water.

Given data:
- Water flow rate = 15 MLD (million liters per day)
- Alum dosage = 35 mg/L
- Alkalinity as CaCO3 = 10 mg/L
- Atomic masses: Al = 27, S = 32, O = 16, H = 1, Ca = 40, C = 12

1. Calculation of Alum dosage per day:
The alum dosage is given as 35 mg/L. As the water flow rate is 15 MLD, we can calculate the alum dosage per day as follows:
Alum dosage = 35 mg/L * 15 MLD = 35 mg/L * 15 * 10^6 L/day = 525 * 10^6 mg/day

2. Calculation of moles of Alum used per day:
To calculate the moles of alum used per day, we need to convert the alum dosage from mg/day to moles/day. The molar mass of Alum (Al2(SO4)3) can be calculated as follows:
Molar mass of Al2(SO4)3 = 2 * Atomic mass of Al + 3 * (Atomic mass of S + 4 * Atomic mass of O)
= 2 * 27 + 3 * (32 + 4 * 16)
= 2 * 27 + 3 * (32 + 64)
= 54 + 3 * 96
= 54 + 288
= 342 g/mol

Now, we can calculate the moles of alum used per day as follows:
Moles of Alum used per day = Alum dosage per day / Molar mass of Al2(SO4)3
= 525 * 10^6 mg/day / 342 g/mol
= (525 * 10^6 mg/day) / (342 * 10^-3 kg/mol)
= (525 * 10^6) / (342 * 10^-3) mol/day
= 1.535 * 10^9 mol/day

3. Calculation of moles of lime required to neutralize the alkalinity:
The alkalinity is given as 10 mg/L. As the water flow rate is 15 MLD, we can calculate the moles of alkalinity per day as follows:
Moles of alkalinity per day = Alkalinity * Water flow rate / Molar mass of CaCO3
= 10 mg/L * 15 * 10^6 L/day / (40 g/mol + 12 g/mol + 3 * 16 g/mol)
= 10 * 10^-3 kg/L * 15 * 10^6 L/day / (40 + 12 + 48) g/mol
= 150 * 10^3 kg/day / 100 g/mol
= 150 * 10^3 / 1000 mol/day
= 150 mol/day

4. Calculation of quantity of quick lime required per year:
To calculate the quantity of quick lime required per year, we need to multiply the moles of lime required per day by the molar mass of CaO and then convert it to tons