How much Ca(OH)2 is required to remove 90.0% of PO4^-3 from 7.5 times 10^6 L of drinking water containing 25 mg/l of PO4^-3?

How much PO4^3- must be removed? That is

0.025 g/L x 7.5E6 L = 187,500 g.

What's 90% of that number?
That is 0.90 x 187,500 = 168,750 g.

How many moles is that?
mol PO4^3- = grams/molar mass PO4^3-

3Ca(OH)2 + 2PO4^3- ==> Ca3(PO4)2 + + 6OH^-
The equation tells you that moles Ca(OH)2 is 3/2 times PO4^3- moles.

I assume the problem wants grams although that isn't shown.
grams Ca(OH)2 = mol Ca(OH)2 x molar mass Ca(OH)2
Check my numbers; you best confirm them.

Well, if we're talking about removing PO4^-3 from drinking water, I hope you're not planning on drinking it straight from the bathtub like a savage. But let's do some math anyway!

To calculate the amount of Ca(OH)2 required, we need to use some stoichiometry. The balanced equation for the reaction between Ca(OH)2 and PO4^-3 is:

3Ca(OH)2 + 2H3PO4 -> Ca3(PO4)2 + 6H2O

From the equation, we can see that 3 moles of Ca(OH)2 are required to react with 2 moles of PO4^-3.

First, let's convert 7.5 times 10^6 L of water to liters.

7.5 times 10^6 L = 7.5 times 10^6 L (since the number won't change)

Now, let's convert the concentration of PO4^-3 from mg/L to moles/L.

25 mg/L x (1 g/1000 mg) x (1 mole/95.97 g) = 2.60018663 times 10^-4 moles/L

Since we want to remove 90% of the PO4^-3, we can multiply the moles/L by 0.9.

2.60018663 times 10^-4 moles/L x 0.9 = 2.34016797 times 10^-4 moles/L

Finally, we can calculate the amount of Ca(OH)2 required.

2.34016797 times 10^-4 moles/L x 7.5 times 10^6 L = 1.75512598 moles

So, approximately 1.76 moles of Ca(OH)2 would be required to remove 90% of the PO4^-3 from 7.5 times 10^6 L of drinking water. But hey, don't forget to consult an actual chemistry expert before you go on a Ca(OH)2 removing spree!

To determine the amount of Ca(OH)2 required to remove 90.0% of PO4^-3 from the drinking water, we can follow these steps:

Step 1: Calculate the initial amount of PO4^-3 in the water.
The initial amount of PO4^-3 in the water can be calculated using the formula: Initial Amount = Volume of water × Concentration of PO4^-3.
Initial Amount = 7.5 × 10^6 L × 25 mg/L = 1.875 × 10^8 mg.

Step 2: Calculate the final amount of PO4^-3 after removal.
The final amount of PO4^-3 remaining in the water after 90.0% removal can be calculated using the formula: Final Amount = Initial Amount × (1 - Removal Percentage).
Final Amount = 1.875 × 10^8 mg × (1 - 0.90) = 1.875 × 10^8 mg × 0.10 = 1.875 × 10^7 mg.

Step 3: Calculate the molar mass of PO4^-3.
The molar mass of PO4^-3 can be calculated by summing the atomic masses of its constituents:
Molar Mass of PO4^-3 = (1 × Atomic Mass of P) + (4 × Atomic Mass of O) = (1 × 31.0 g/mol) + (4 × 16.0 g/mol) = 31.0 g/mol + 64.0 g/mol = 95.0 g/mol.

Step 4: Calculate the moles of PO4^-3 remaining in the water.
The moles of PO4^-3 remaining in the water can be calculated using the formula: Moles of PO4^-3 = Final Amount / Molar Mass of PO4^-3.
Moles of PO4^-3 = (1.875 × 10^7 mg) / (95.0 g/mol) = 1.97 × 10^5 mol.

Step 5: Determine the moles of Ca(OH)2 required for removal.
The balanced chemical equation for the reaction between Ca(OH)2 and PO4^-3 is:
3 Ca(OH)2 + 2 H3PO4 → Ca3(PO4)2 + 6 H2O.

From the equation, we can see that 3 moles of Ca(OH)2 react with 2 moles of PO4^-3.
Therefore, the moles of Ca(OH)2 required for removal can be calculated using the formula:
Moles of Ca(OH)2 = (2 moles of PO4^-3) × (3 moles of Ca(OH)2 / 2 moles of PO4^-3) = 3 moles of Ca(OH)2.

Step 6: Convert moles of Ca(OH)2 to grams.
The molar mass of Ca(OH)2 = (1 × Atomic Mass of Ca) + (2 × Atomic Mass of O) + (2 × Atomic Mass of H) = (1 × 40.1 g/mol) + (2 × 16.0 g/mol) + (2 × 1.0 g/mol) = 40.1 g/mol + 32.0 g/mol + 2.0 g/mol = 74.1 g/mol.

The grams of Ca(OH)2 required can be calculated using the formula: Grams of Ca(OH)2 = Moles of Ca(OH)2 × Molar Mass of Ca(OH)2.
Grams of Ca(OH)2 = (3 moles) × (74.1 g/mol) = 222.3 g.

Therefore, approximately 222.3 grams of Ca(OH)2 is required to remove 90.0% of PO4^-3 from 7.5 × 10^6 L of drinking water containing 25 mg/L of PO4^-3.

To calculate the amount of Ca(OH)2 required to remove 90.0% of PO4^-3 from the drinking water, we need to follow a few steps:

Step 1: Calculate the initial amount of PO4^-3 in the water.
To do this, we multiply the volume of the water (7.5 times 10^6 L) by the concentration of PO4^-3 (25 mg/L):

Initial amount of PO4^-3 = 7.5 times 10^6 L × 25 mg/L

Step 2: Calculate the final amount of PO4^-3 remaining in the water.
Since we want to remove 90.0% of the PO4^-3, the final amount of PO4^-3 can be calculated as:

Final amount of PO4^-3 = (Initial amount of PO4^-3) × (1 - 0.90)

Step 3: Calculate the amount of PO4^-3 that needs to be removed.
The amount of PO4^-3 that needs to be removed can be calculated by subtracting the final amount of PO4^-3 from the initial amount:

Amount of PO4^-3 to remove = Initial amount of PO4^-3 - Final amount of PO4^-3

Step 4: Calculate the amount of Ca(OH)2 required.
The reaction between Ca(OH)2 and PO4^-3 is as follows:

Ca(OH)2 + PO4^-3 → CaPO4 + 2OH^-

From the reaction, we can see that one mole of Ca(OH)2 reacts with one mole of PO4^-3. Therefore, to remove the calculated amount of PO4^-3 (in moles), we need the same amount of Ca(OH)2 (in moles).

Amount of Ca(OH)2 required (in moles) = Amount of PO4^-3 to remove (in moles)

Finally, you can convert the moles of Ca(OH)2 required to grams by multiplying it by the molar mass of Ca(OH)2.

Note:
Make sure to consider the molar mass of PO4^-3 when calculating the amounts in moles.

Hope this helps!