To calculate the amount of calcium hydroxide $\text{Ca(OH)}_2$ dosed per year in kiloton (kton), we'll break it down step by step.

Given:

• Volume of water treated per year: $21,000,000 \, \text{m}^3 \, \text{per year}$
• Concentration of $\text{Ca(OH)}_2$ dose: $1.14 \, \text{mmol/L}$
• Molar mass of $\text{Ca(OH)}_2$: $74 \, \text{g/mol}$

Step 1: Convert volume of water to liters

Since 1 cubic meter (m$^3$) = 1000 liters (L), we have:

$21,000,000 \, \text{m}^3 \times 1000 = 21,000,000,000 \, \text{L}$

Step 2: Convert concentration to moles of $\text{Ca(OH)}_2$

Given the concentration of $\text{Ca(OH)}_2$ is $1.14 \, \text{mmol/L}$, we first convert this to mol/L:

$1.14 \, \text{mmol/L} = 1.14 \times 10^{-3} \, \text{mol/L}$

Now, multiply this concentration by the total volume of water in liters to find the total moles of $\text{Ca(OH)}_2$:

$1.14 \times 10^{-3} \, \text{mol/L} \times 21,000,000,000 \, \text{L} = 23,940,000 \, \text{mol}$

Step 3: Convert moles to mass

Now, using the molar mass of $\text{Ca(OH)}_2$, which is $74 \, \text{g/mol}$, we can convert moles to grams:

$23,940,000 \, \text{mol} \times 74 \, \text{g/mol} = 1,770,360,000 \, \text{g}$

Convert grams to kilograms (since 1 kg = 1000 g):

$1,770,360,000 \, \text{g} = 1,770,360 \, \text{kg}$

Step 4: Convert kilograms to kiloton

Since 1 kiloton (kton) = 1,000,000 kg, we convert:

$1,770,360 \, \text{kg} = 1.77 \, \text{kton}$

Final Answer:

The amount of $\text{Ca(OH)}_2$ dosed per year is 2 kton (rounded to 0 decimals).

Would you like further details or have any questions?

Here are 5 related questions:

1. What are other chemicals used in water treatment and their dosages?
2. How does calcium hydroxide soften water?
3. How do you calculate the molar mass of a compound?
4. What is the significance of millimolar (mmol/L) in chemical dosing?
5. How is the efficiency of water treatment plants measured?

Tip: Always double-check unit conversions when working with large volumes and dosages.