phosphorus removal chemical dose calculation using metal to phosphorus ratio Given metal to phosphorus molar ratio (MeP) = 1.6000; phosphorus mass to remove (M_P) = 44.0000 kg/day, determine the metal salt dose mass (M_Me) in kg/day.

Start with the thinking

• The governing relation printed in this handbook section is Phosphorus Removal Equations.
• Everything except M_Me is given, so isolate M_Me symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Phosphorus removal equations determine the metal salt dose needed for chemical precipitation of phosphorus.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

2. Step 2 — Rearrange symbolically for M_Me:

   $$M_{Me} = \left(\dfrac{Me}{P}\right)M_P$$

3. Step 3 — List the givens: metal to phosphorus molar ratio (MeP) = 1.6000, phosphorus mass to remove (M_P) = 44.0000 kg/day.

4. Step 4 — Substitute the given values:

   $$M_{Me} = \left(\dfrac{Me}{P}\right)M_P$$

5. Step 5 — Evaluate:

   $$M_{Me} = 70.4000\ \text{kg/day}$$

6. Step 6 — Check: returning M_Me = 70.4000 kg/day to

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 140.8 — kept a factor of two that cancels in the correct rearrangement.
• 35.2000 — dropped that same factor in the other direction.
• 77.4400 — rounded an intermediate value before the final step.

Reference: FE Handbook — Phosphorus Removal Equations

phosphorus removal equations for alum dosing at a treatment plant Given metal to phosphorus molar ratio (MeP) = 1.7000; metal salt dose mass (M_Me) = 971.0 kg/day, determine the phosphorus mass to remove (M_P) in kg/day.

Start with the thinking

• The governing relation printed in this handbook section is Phosphorus Removal Equations.
• Everything except M_P is given, so isolate M_P symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Phosphorus removal equations determine the metal salt dose needed for chemical precipitation of phosphorus.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

2. Step 2 — Rearrange symbolically for M_P:

   $$M_{P} = \dfrac{M_{Me}}{Me/P}$$

3. Step 3 — List the givens: metal to phosphorus molar ratio (MeP) = 1.7000, metal salt dose mass (M_Me) = 971.0 kg/day.

4. Step 4 — Substitute the given values:

   $$M_{P} = \dfrac{971.0}{Me/P}$$

5. Step 5 — Evaluate:

   $$M_{P} = 571.2\ \text{kg/day}$$

6. Step 6 — Check: returning M_P = 571.2 kg/day to

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1,142 — kept a factor of two that cancels in the correct rearrangement.
• 285.6 — dropped that same factor in the other direction.
• 628.3 — rounded an intermediate value before the final step.

Reference: FE Handbook — Phosphorus Removal Equations

phosphorus removal via chemical precipitation dose determination Given phosphorus mass to remove (M_P) = 155.0 kg/day; metal salt dose mass (M_Me) = 605.0 kg/day, determine the metal to phosphorus molar ratio (MeP).

Start with the thinking

• The governing relation printed in this handbook section is Phosphorus Removal Equations.
• Everything except MeP is given, so isolate MeP symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Phosphorus removal equations determine the metal salt dose needed for chemical precipitation of phosphorus.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

2. Step 2 — Rearrange symbolically for MeP:

   $$MeP = \dfrac{M_{Me}}{M_P}$$

3. Step 3 — List the givens: phosphorus mass to remove (M_P) = 155.0 kg/day, metal salt dose mass (M_Me) = 605.0 kg/day.

4. Step 4 — Substitute the given values:

   $$MeP = \dfrac{605.0}{M_P}$$

5. Step 5 — Evaluate:

   $$MeP = 3.9032$$

6. Step 6 — Check: returning MeP = 3.9032 to

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 7.8065 — kept a factor of two that cancels in the correct rearrangement.
• 1.9516 — dropped that same factor in the other direction.
• 4.2935 — rounded an intermediate value before the final step.

Reference: FE Handbook — Phosphorus Removal Equations

phosphorus removal chemical dose calculation using metal to phosphorus ratio Given metal to phosphorus molar ratio (MeP) = 1.5500; phosphorus mass to remove (M_P) = 100.0 kg/day, determine the metal salt dose mass (M_Me) in kg/day.

Start with the thinking

• The governing relation printed in this handbook section is Phosphorus Removal Equations.
• Everything except M_Me is given, so isolate M_Me symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Phosphorus removal equations determine the metal salt dose needed for chemical precipitation of phosphorus.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

2. Step 2 — Rearrange symbolically for M_Me:

   $$M_{Me} = \left(\dfrac{Me}{P}\right)M_P$$

3. Step 3 — List the givens: metal to phosphorus molar ratio (MeP) = 1.5500, phosphorus mass to remove (M_P) = 100.0 kg/day.

4. Step 4 — Substitute the given values:

   $$M_{Me} = \left(\dfrac{Me}{P}\right)M_P$$

5. Step 5 — Evaluate:

   $$M_{Me} = 155.0\ \text{kg/day}$$

6. Step 6 — Check: returning M_Me = 155.0 kg/day to

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 310.0 — kept a factor of two that cancels in the correct rearrangement.
• 77.5000 — dropped that same factor in the other direction.
• 170.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Phosphorus Removal Equations

phosphorus removal equations for alum dosing at a treatment plant Given metal to phosphorus molar ratio (MeP) = 2.7000; metal salt dose mass (M_Me) = 448.0 kg/day, determine the phosphorus mass to remove (M_P) in kg/day.

Start with the thinking

• The governing relation printed in this handbook section is Phosphorus Removal Equations.
• Everything except M_P is given, so isolate M_P symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Phosphorus removal equations determine the metal salt dose needed for chemical precipitation of phosphorus.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

2. Step 2 — Rearrange symbolically for M_P:

   $$M_{P} = \dfrac{M_{Me}}{Me/P}$$

3. Step 3 — List the givens: metal to phosphorus molar ratio (MeP) = 2.7000, metal salt dose mass (M_Me) = 448.0 kg/day.

4. Step 4 — Substitute the given values:

   $$M_{P} = \dfrac{448.0}{Me/P}$$

5. Step 5 — Evaluate:

   $$M_{P} = 165.9\ \text{kg/day}$$

6. Step 6 — Check: returning M_P = 165.9 kg/day to

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 331.9 — kept a factor of two that cancels in the correct rearrangement.
• 82.9630 — dropped that same factor in the other direction.
• 182.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Phosphorus Removal Equations

phosphorus removal via chemical precipitation dose determination Given phosphorus mass to remove (M_P) = 355.0 kg/day; metal salt dose mass (M_Me) = 608.0 kg/day, determine the metal to phosphorus molar ratio (MeP).

Start with the thinking

• The governing relation printed in this handbook section is Phosphorus Removal Equations.
• Everything except MeP is given, so isolate MeP symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Phosphorus removal equations determine the metal salt dose needed for chemical precipitation of phosphorus.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

2. Step 2 — Rearrange symbolically for MeP:

   $$MeP = \dfrac{M_{Me}}{M_P}$$

3. Step 3 — List the givens: phosphorus mass to remove (M_P) = 355.0 kg/day, metal salt dose mass (M_Me) = 608.0 kg/day.

4. Step 4 — Substitute the given values:

   $$MeP = \dfrac{608.0}{M_P}$$

5. Step 5 — Evaluate:

   $$MeP = 1.7127$$

6. Step 6 — Check: returning MeP = 1.7127 to

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 3.4254 — kept a factor of two that cancels in the correct rearrangement.
• 0.8563 — dropped that same factor in the other direction.
• 1.8839 — rounded an intermediate value before the final step.

Reference: FE Handbook — Phosphorus Removal Equations

phosphorus removal chemical dose calculation using metal to phosphorus ratio Given metal to phosphorus molar ratio (MeP) = 1.9500; phosphorus mass to remove (M_P) = 121.0 kg/day, determine the metal salt dose mass (M_Me) in kg/day.

Start with the thinking

• The governing relation printed in this handbook section is Phosphorus Removal Equations.
• Everything except M_Me is given, so isolate M_Me symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Phosphorus removal equations determine the metal salt dose needed for chemical precipitation of phosphorus.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

2. Step 2 — Rearrange symbolically for M_Me:

   $$M_{Me} = \left(\dfrac{Me}{P}\right)M_P$$

3. Step 3 — List the givens: metal to phosphorus molar ratio (MeP) = 1.9500, phosphorus mass to remove (M_P) = 121.0 kg/day.

4. Step 4 — Substitute the given values:

   $$M_{Me} = \left(\dfrac{Me}{P}\right)M_P$$

5. Step 5 — Evaluate:

   $$M_{Me} = 236.0\ \text{kg/day}$$

6. Step 6 — Check: returning M_Me = 236.0 kg/day to

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 471.9 — kept a factor of two that cancels in the correct rearrangement.
• 118.0 — dropped that same factor in the other direction.
• 259.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Phosphorus Removal Equations

phosphorus removal equations for alum dosing at a treatment plant Given metal to phosphorus molar ratio (MeP) = 1.9000; metal salt dose mass (M_Me) = 218.0 kg/day, determine the phosphorus mass to remove (M_P) in kg/day.

Start with the thinking

• The governing relation printed in this handbook section is Phosphorus Removal Equations.
• Everything except M_P is given, so isolate M_P symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Phosphorus removal equations determine the metal salt dose needed for chemical precipitation of phosphorus.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

2. Step 2 — Rearrange symbolically for M_P:

   $$M_{P} = \dfrac{M_{Me}}{Me/P}$$

3. Step 3 — List the givens: metal to phosphorus molar ratio (MeP) = 1.9000, metal salt dose mass (M_Me) = 218.0 kg/day.

4. Step 4 — Substitute the given values:

   $$M_{P} = \dfrac{218.0}{Me/P}$$

5. Step 5 — Evaluate:

   $$M_{P} = 114.7\ \text{kg/day}$$

6. Step 6 — Check: returning M_P = 114.7 kg/day to

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 229.5 — kept a factor of two that cancels in the correct rearrangement.
• 57.3684 — dropped that same factor in the other direction.
• 126.2 — rounded an intermediate value before the final step.

Reference: FE Handbook — Phosphorus Removal Equations

phosphorus removal via chemical precipitation dose determination Given phosphorus mass to remove (M_P) = 353.0 kg/day; metal salt dose mass (M_Me) = 120.0 kg/day, determine the metal to phosphorus molar ratio (MeP).

Start with the thinking

• The governing relation printed in this handbook section is Phosphorus Removal Equations.
• Everything except MeP is given, so isolate MeP symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Phosphorus removal equations determine the metal salt dose needed for chemical precipitation of phosphorus.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

2. Step 2 — Rearrange symbolically for MeP:

   $$MeP = \dfrac{M_{Me}}{M_P}$$

3. Step 3 — List the givens: phosphorus mass to remove (M_P) = 353.0 kg/day, metal salt dose mass (M_Me) = 120.0 kg/day.

4. Step 4 — Substitute the given values:

   $$MeP = \dfrac{120.0}{M_P}$$

5. Step 5 — Evaluate:

   $$MeP = 0.3399$$

6. Step 6 — Check: returning MeP = 0.3399 to

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.6799 — kept a factor of two that cancels in the correct rearrangement.
• 0.1700 — dropped that same factor in the other direction.
• 0.3739 — rounded an intermediate value before the final step.

Reference: FE Handbook — Phosphorus Removal Equations

phosphorus removal chemical dose calculation using metal to phosphorus ratio Given metal to phosphorus molar ratio (MeP) = 1.4500; phosphorus mass to remove (M_P) = 285.0 kg/day, determine the metal salt dose mass (M_Me) in kg/day.

Start with the thinking

• The governing relation printed in this handbook section is Phosphorus Removal Equations.
• Everything except M_Me is given, so isolate M_Me symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Phosphorus removal equations determine the metal salt dose needed for chemical precipitation of phosphorus.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

2. Step 2 — Rearrange symbolically for M_Me:

   $$M_{Me} = \left(\dfrac{Me}{P}\right)M_P$$

3. Step 3 — List the givens: metal to phosphorus molar ratio (MeP) = 1.4500, phosphorus mass to remove (M_P) = 285.0 kg/day.

4. Step 4 — Substitute the given values:

   $$M_{Me} = \left(\dfrac{Me}{P}\right)M_P$$

5. Step 5 — Evaluate:

   $$M_{Me} = 413.3\ \text{kg/day}$$

6. Step 6 — Check: returning M_Me = 413.3 kg/day to

   $$M_{Me} = \dfrac{Me}{P} \times M_P$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 826.5 — kept a factor of two that cancels in the correct rearrangement.
• 206.6 — dropped that same factor in the other direction.
• 454.6 — rounded an intermediate value before the final step.

Reference: FE Handbook — Phosphorus Removal Equations