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How to convert P2O5 concentration to H3PO4 concentration?
Concentration InquiryElectronic oxygen concentration sensorHow can I convert the number of particles per liter to gram per literCan I convert from an emission amount (in tonnes per area) to a concentration (in ppbv)?How does dissolving a hydrate affect concentration?How to convert ppm to g/m3?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I'm confused why $ceP2O5$ concentrations are reported for materials which have no $ceP2O5$ in them. I guess it simplifies things once you understand it. However, at the moment, I don't understand how the conversion is done.
For example. the Wikipedia page on phosphoric acid states that a $ceP2O5$ concentration of 70% is equivalent to nearly 100% $ceH3PO4$. Can someone explain this conversion?
concentration organophosphorus-compounds
edited 7 hours ago
M. Farooq
6,4128 silver badges27 bronze badges
asked 9 hours ago
dlightdlight
242 bronze badges
$endgroup$
add a comment |
$begingroup$
I'm confused why $ceP2O5$ concentrations are reported for materials which have no $ceP2O5$ in them. I guess it simplifies things once you understand it. However, at the moment, I don't understand how the conversion is done.
For example. the Wikipedia page on phosphoric acid states that a $ceP2O5$ concentration of 70% is equivalent to nearly 100% $ceH3PO4$. Can someone explain this conversion?
concentration organophosphorus-compounds
edited 7 hours ago
M. Farooq
6,4128 silver badges27 bronze badges
asked 9 hours ago
dlightdlight
242 bronze badges
$endgroup$
1
$begingroup$
What's to explain? P2O5 reacts with water. Do you know what the product is? Can you balance the reaction?
$endgroup$
– Ivan Neretin
9 hours ago
$begingroup$
okay I didnt realize the concentration was in reference to water because its often discussed as purity etc. makes sense thanks.
$endgroup$
– dlight
9 hours ago
$begingroup$
@IvanNeretin Im reading a paper which states energy used to process phosphoric acid is "6500 kwh to make a ton of phosphoric acid (P2O5)". I want to express this in kWh/ton H3PO4. So I get 4710 kWh/ton H3PO4. Agreed? The paper is: A Technical Review of the Improved Hard Process
$endgroup$
– dlight
8 hours ago
$begingroup$
It is a common practice to express content of the component A in the equivalent content of the component B. There is an older unit of the water "hardness" dGH ( degree of General/German Hardness ) where 1 dGH means equivalent of 10 mg CaO / 1 L. But there is no CaO in water. Similarly, in fertilizers, there is AFAIK historical custom to use content of oxides, even if there are none.
$endgroup$
– Poutnik
8 hours ago
1
$begingroup$
P2O5 + 3H2O --> 2H3PO4. Using molecular weights P2O5/2H3PO4 = 142/198 = 0.71 = 71%
$endgroup$
– user55119
3 hours ago
add a comment |
$begingroup$
I'm confused why $ceP2O5$ concentrations are reported for materials which have no $ceP2O5$ in them. I guess it simplifies things once you understand it. However, at the moment, I don't understand how the conversion is done.
For example. the Wikipedia page on phosphoric acid states that a $ceP2O5$ concentration of 70% is equivalent to nearly 100% $ceH3PO4$. Can someone explain this conversion?
concentration organophosphorus-compounds
edited 7 hours ago
M. Farooq
6,4128 silver badges27 bronze badges
asked 9 hours ago
dlightdlight
242 bronze badges
$endgroup$
I'm confused why $ceP2O5$ concentrations are reported for materials which have no $ceP2O5$ in them. I guess it simplifies things once you understand it. However, at the moment, I don't understand how the conversion is done.
For example. the Wikipedia page on phosphoric acid states that a $ceP2O5$ concentration of 70% is equivalent to nearly 100% $ceH3PO4$. Can someone explain this conversion?
concentration organophosphorus-compounds
concentration organophosphorus-compounds
edited 7 hours ago
M. Farooq
6,4128 silver badges27 bronze badges
asked 9 hours ago
dlightdlight
242 bronze badges
edited 7 hours ago
M. Farooq
6,4128 silver badges27 bronze badges
asked 9 hours ago
dlightdlight
242 bronze badges
edited 7 hours ago
M. Farooq
6,4128 silver badges27 bronze badges
edited 7 hours ago
M. Farooq
6,4128 silver badges27 bronze badges
edited 7 hours ago
M. Farooq
6,4128 silver badges27 bronze badges
6,4128 silver badges27 bronze badges
asked 9 hours ago
dlightdlight
242 bronze badges
asked 9 hours ago
dlightdlight
242 bronze badges
asked 9 hours ago
dlightdlight
242 bronze badges
242 bronze badges
1
$begingroup$
What's to explain? P2O5 reacts with water. Do you know what the product is? Can you balance the reaction?
$endgroup$
– Ivan Neretin
9 hours ago
$begingroup$
okay I didnt realize the concentration was in reference to water because its often discussed as purity etc. makes sense thanks.
$endgroup$
– dlight
9 hours ago
$begingroup$
@IvanNeretin Im reading a paper which states energy used to process phosphoric acid is "6500 kwh to make a ton of phosphoric acid (P2O5)". I want to express this in kWh/ton H3PO4. So I get 4710 kWh/ton H3PO4. Agreed? The paper is: A Technical Review of the Improved Hard Process
$endgroup$
– dlight
8 hours ago
$begingroup$
It is a common practice to express content of the component A in the equivalent content of the component B. There is an older unit of the water "hardness" dGH ( degree of General/German Hardness ) where 1 dGH means equivalent of 10 mg CaO / 1 L. But there is no CaO in water. Similarly, in fertilizers, there is AFAIK historical custom to use content of oxides, even if there are none.
$endgroup$
– Poutnik
8 hours ago
1
$begingroup$
P2O5 + 3H2O --> 2H3PO4. Using molecular weights P2O5/2H3PO4 = 142/198 = 0.71 = 71%
$endgroup$
– user55119
3 hours ago
add a comment |
1
$begingroup$
What's to explain? P2O5 reacts with water. Do you know what the product is? Can you balance the reaction?
$endgroup$
– Ivan Neretin
9 hours ago
$begingroup$
okay I didnt realize the concentration was in reference to water because its often discussed as purity etc. makes sense thanks.
$endgroup$
– dlight
9 hours ago
$begingroup$
@IvanNeretin Im reading a paper which states energy used to process phosphoric acid is "6500 kwh to make a ton of phosphoric acid (P2O5)". I want to express this in kWh/ton H3PO4. So I get 4710 kWh/ton H3PO4. Agreed? The paper is: A Technical Review of the Improved Hard Process
$endgroup$
– dlight
8 hours ago
$begingroup$
It is a common practice to express content of the component A in the equivalent content of the component B. There is an older unit of the water "hardness" dGH ( degree of General/German Hardness ) where 1 dGH means equivalent of 10 mg CaO / 1 L. But there is no CaO in water. Similarly, in fertilizers, there is AFAIK historical custom to use content of oxides, even if there are none.
$endgroup$
– Poutnik
8 hours ago
1
$begingroup$
P2O5 + 3H2O --> 2H3PO4. Using molecular weights P2O5/2H3PO4 = 142/198 = 0.71 = 71%
$endgroup$
– user55119
3 hours ago
1
1
$begingroup$
What's to explain? P2O5 reacts with water. Do you know what the product is? Can you balance the reaction?
$endgroup$
– Ivan Neretin
9 hours ago
$begingroup$
What's to explain? P2O5 reacts with water. Do you know what the product is? Can you balance the reaction?
$endgroup$
– Ivan Neretin
9 hours ago
$begingroup$
okay I didnt realize the concentration was in reference to water because its often discussed as purity etc. makes sense thanks.
$endgroup$
– dlight
9 hours ago
$begingroup$
okay I didnt realize the concentration was in reference to water because its often discussed as purity etc. makes sense thanks.
$endgroup$
– dlight
9 hours ago
$begingroup$
@IvanNeretin Im reading a paper which states energy used to process phosphoric acid is "6500 kwh to make a ton of phosphoric acid (P2O5)". I want to express this in kWh/ton H3PO4. So I get 4710 kWh/ton H3PO4. Agreed? The paper is: A Technical Review of the Improved Hard Process
$endgroup$
– dlight
8 hours ago
$begingroup$
@IvanNeretin Im reading a paper which states energy used to process phosphoric acid is "6500 kwh to make a ton of phosphoric acid (P2O5)". I want to express this in kWh/ton H3PO4. So I get 4710 kWh/ton H3PO4. Agreed? The paper is: A Technical Review of the Improved Hard Process
$endgroup$
– dlight
8 hours ago
$begingroup$
It is a common practice to express content of the component A in the equivalent content of the component B. There is an older unit of the water "hardness" dGH ( degree of General/German Hardness ) where 1 dGH means equivalent of 10 mg CaO / 1 L. But there is no CaO in water. Similarly, in fertilizers, there is AFAIK historical custom to use content of oxides, even if there are none.
$endgroup$
– Poutnik
8 hours ago
$begingroup$
It is a common practice to express content of the component A in the equivalent content of the component B. There is an older unit of the water "hardness" dGH ( degree of General/German Hardness ) where 1 dGH means equivalent of 10 mg CaO / 1 L. But there is no CaO in water. Similarly, in fertilizers, there is AFAIK historical custom to use content of oxides, even if there are none.
$endgroup$
– Poutnik
8 hours ago
1
1
$begingroup$
P2O5 + 3H2O --> 2H3PO4. Using molecular weights P2O5/2H3PO4 = 142/198 = 0.71 = 71%
$endgroup$
– user55119
3 hours ago
$begingroup$
P2O5 + 3H2O --> 2H3PO4. Using molecular weights P2O5/2H3PO4 = 142/198 = 0.71 = 71%
$endgroup$
– user55119
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your question is a valid analytical chemistry question which has deep roots in history. No the reason is not that $ceP2O5$ reacts with water to from phosphoric acid and that is why $ceP2O5$ values are quoted. This is a very old historical tradition. In older analytical chemistry, chemists had only two main tools to analyse something namely, titration and gravimetry. Oxygen was also used to determine atomic weights, and people were interested in the ratios of element:oxygen ratio. One person even got a Nobel Prize as well. Generally, when the precipitates were heated, what was left was oxides. In really old books (at least 100 year old) you will notice that many analyses were quoted as oxides. For example, this is true in fertilizer industry to quote potassium as $ceK2O$ and so on. Now if you strongly heat a precipitate phosphate, one will not get a $ceP2O5$ but the convention to report elements as oxides continues.
Second part: I will let you solve the rest of the problem: Start from here
1 mol $ceP2O5$ = 2 mol $ceH3PO4$
and do wt percentage conversions to moles. A 70% $ceP2O5$ solution in water means 70 g $ceP2O5$ is present in 100 grams of solution. Do mol conversions. You will also need the density of the solution.
answered 8 hours ago
M. FarooqM. Farooq
6,4128 silver badges27 bronze badges
$endgroup$
$begingroup$
Thanks for the answer I believe I solved it based on first poster response as follows (in 1 kg). 700g P2O5 x 1 mol/ 142 g = 4.9 mol P2O5. 300 g H2O is 16.7 mol H2O. 4.9 mol P2O5 + 17.78 mol H2O --> 9.8 mol H3PO4 Mass of 9.8 mol H3PO4 * 98 g mol = 960.4 g H3PO4
$endgroup$
– dlight
7 hours ago
$begingroup$
Seems right, when dealing with percentage concentrations, it is better to use 100g as the unit. So using your numbers, it is 96% w/w phosphoric acid.
$endgroup$
– M. Farooq
7 hours ago
add a comment |
$begingroup$
$ceP2O5$ content is useful as it is the phosphorous content that a scientist often cares about when using phosphoric acids. This might seem silly given that as you pointed out anything from $0-70% ceP2O5$ could just be expressed as a percentage of $ceH3PO4$ but there are times when a higher $ceP2O5$ concentration is desired. For example polyphosphoric acids are useful reagents that have more phosphorous content that phosphoric acid. It would be ambiguous to state the concentration as $115% ceH3PO4$ (what does 115% mean) so instead procedures are reported in terms of $ceP2O5$ concentration with the balance assumed to be water (for those in the know).
answered 8 hours ago
A.K.A.K.
10.8k6 gold badges31 silver badges75 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your question is a valid analytical chemistry question which has deep roots in history. No the reason is not that $ceP2O5$ reacts with water to from phosphoric acid and that is why $ceP2O5$ values are quoted. This is a very old historical tradition. In older analytical chemistry, chemists had only two main tools to analyse something namely, titration and gravimetry. Oxygen was also used to determine atomic weights, and people were interested in the ratios of element:oxygen ratio. One person even got a Nobel Prize as well. Generally, when the precipitates were heated, what was left was oxides. In really old books (at least 100 year old) you will notice that many analyses were quoted as oxides. For example, this is true in fertilizer industry to quote potassium as $ceK2O$ and so on. Now if you strongly heat a precipitate phosphate, one will not get a $ceP2O5$ but the convention to report elements as oxides continues.
Second part: I will let you solve the rest of the problem: Start from here
1 mol $ceP2O5$ = 2 mol $ceH3PO4$
and do wt percentage conversions to moles. A 70% $ceP2O5$ solution in water means 70 g $ceP2O5$ is present in 100 grams of solution. Do mol conversions. You will also need the density of the solution.
answered 8 hours ago
M. FarooqM. Farooq
6,4128 silver badges27 bronze badges
$endgroup$
$begingroup$
Thanks for the answer I believe I solved it based on first poster response as follows (in 1 kg). 700g P2O5 x 1 mol/ 142 g = 4.9 mol P2O5. 300 g H2O is 16.7 mol H2O. 4.9 mol P2O5 + 17.78 mol H2O --> 9.8 mol H3PO4 Mass of 9.8 mol H3PO4 * 98 g mol = 960.4 g H3PO4
$endgroup$
– dlight
7 hours ago
$begingroup$
Seems right, when dealing with percentage concentrations, it is better to use 100g as the unit. So using your numbers, it is 96% w/w phosphoric acid.
$endgroup$
– M. Farooq
7 hours ago
add a comment |
$begingroup$
Your question is a valid analytical chemistry question which has deep roots in history. No the reason is not that $ceP2O5$ reacts with water to from phosphoric acid and that is why $ceP2O5$ values are quoted. This is a very old historical tradition. In older analytical chemistry, chemists had only two main tools to analyse something namely, titration and gravimetry. Oxygen was also used to determine atomic weights, and people were interested in the ratios of element:oxygen ratio. One person even got a Nobel Prize as well. Generally, when the precipitates were heated, what was left was oxides. In really old books (at least 100 year old) you will notice that many analyses were quoted as oxides. For example, this is true in fertilizer industry to quote potassium as $ceK2O$ and so on. Now if you strongly heat a precipitate phosphate, one will not get a $ceP2O5$ but the convention to report elements as oxides continues.
Second part: I will let you solve the rest of the problem: Start from here
1 mol $ceP2O5$ = 2 mol $ceH3PO4$
and do wt percentage conversions to moles. A 70% $ceP2O5$ solution in water means 70 g $ceP2O5$ is present in 100 grams of solution. Do mol conversions. You will also need the density of the solution.
answered 8 hours ago
M. FarooqM. Farooq
6,4128 silver badges27 bronze badges
$endgroup$
$begingroup$
Thanks for the answer I believe I solved it based on first poster response as follows (in 1 kg). 700g P2O5 x 1 mol/ 142 g = 4.9 mol P2O5. 300 g H2O is 16.7 mol H2O. 4.9 mol P2O5 + 17.78 mol H2O --> 9.8 mol H3PO4 Mass of 9.8 mol H3PO4 * 98 g mol = 960.4 g H3PO4
$endgroup$
– dlight
7 hours ago
$begingroup$
Seems right, when dealing with percentage concentrations, it is better to use 100g as the unit. So using your numbers, it is 96% w/w phosphoric acid.
$endgroup$
– M. Farooq
7 hours ago
add a comment |
$begingroup$
Your question is a valid analytical chemistry question which has deep roots in history. No the reason is not that $ceP2O5$ reacts with water to from phosphoric acid and that is why $ceP2O5$ values are quoted. This is a very old historical tradition. In older analytical chemistry, chemists had only two main tools to analyse something namely, titration and gravimetry. Oxygen was also used to determine atomic weights, and people were interested in the ratios of element:oxygen ratio. One person even got a Nobel Prize as well. Generally, when the precipitates were heated, what was left was oxides. In really old books (at least 100 year old) you will notice that many analyses were quoted as oxides. For example, this is true in fertilizer industry to quote potassium as $ceK2O$ and so on. Now if you strongly heat a precipitate phosphate, one will not get a $ceP2O5$ but the convention to report elements as oxides continues.
Second part: I will let you solve the rest of the problem: Start from here
1 mol $ceP2O5$ = 2 mol $ceH3PO4$
and do wt percentage conversions to moles. A 70% $ceP2O5$ solution in water means 70 g $ceP2O5$ is present in 100 grams of solution. Do mol conversions. You will also need the density of the solution.
answered 8 hours ago
M. FarooqM. Farooq
6,4128 silver badges27 bronze badges
$endgroup$
Your question is a valid analytical chemistry question which has deep roots in history. No the reason is not that $ceP2O5$ reacts with water to from phosphoric acid and that is why $ceP2O5$ values are quoted. This is a very old historical tradition. In older analytical chemistry, chemists had only two main tools to analyse something namely, titration and gravimetry. Oxygen was also used to determine atomic weights, and people were interested in the ratios of element:oxygen ratio. One person even got a Nobel Prize as well. Generally, when the precipitates were heated, what was left was oxides. In really old books (at least 100 year old) you will notice that many analyses were quoted as oxides. For example, this is true in fertilizer industry to quote potassium as $ceK2O$ and so on. Now if you strongly heat a precipitate phosphate, one will not get a $ceP2O5$ but the convention to report elements as oxides continues.
Second part: I will let you solve the rest of the problem: Start from here
1 mol $ceP2O5$ = 2 mol $ceH3PO4$
and do wt percentage conversions to moles. A 70% $ceP2O5$ solution in water means 70 g $ceP2O5$ is present in 100 grams of solution. Do mol conversions. You will also need the density of the solution.
answered 8 hours ago
M. FarooqM. Farooq
6,4128 silver badges27 bronze badges
answered 8 hours ago
M. FarooqM. Farooq
6,4128 silver badges27 bronze badges
answered 8 hours ago
M. FarooqM. Farooq
6,4128 silver badges27 bronze badges
answered 8 hours ago
M. FarooqM. Farooq
6,4128 silver badges27 bronze badges
6,4128 silver badges27 bronze badges
$begingroup$
Thanks for the answer I believe I solved it based on first poster response as follows (in 1 kg). 700g P2O5 x 1 mol/ 142 g = 4.9 mol P2O5. 300 g H2O is 16.7 mol H2O. 4.9 mol P2O5 + 17.78 mol H2O --> 9.8 mol H3PO4 Mass of 9.8 mol H3PO4 * 98 g mol = 960.4 g H3PO4
$endgroup$
– dlight
7 hours ago
$begingroup$
Seems right, when dealing with percentage concentrations, it is better to use 100g as the unit. So using your numbers, it is 96% w/w phosphoric acid.
$endgroup$
– M. Farooq
7 hours ago
add a comment |
$begingroup$
Thanks for the answer I believe I solved it based on first poster response as follows (in 1 kg). 700g P2O5 x 1 mol/ 142 g = 4.9 mol P2O5. 300 g H2O is 16.7 mol H2O. 4.9 mol P2O5 + 17.78 mol H2O --> 9.8 mol H3PO4 Mass of 9.8 mol H3PO4 * 98 g mol = 960.4 g H3PO4
$endgroup$
– dlight
7 hours ago
$begingroup$
Seems right, when dealing with percentage concentrations, it is better to use 100g as the unit. So using your numbers, it is 96% w/w phosphoric acid.
$endgroup$
– M. Farooq
7 hours ago
$begingroup$
Thanks for the answer I believe I solved it based on first poster response as follows (in 1 kg). 700g P2O5 x 1 mol/ 142 g = 4.9 mol P2O5. 300 g H2O is 16.7 mol H2O. 4.9 mol P2O5 + 17.78 mol H2O --> 9.8 mol H3PO4 Mass of 9.8 mol H3PO4 * 98 g mol = 960.4 g H3PO4
$endgroup$
– dlight
7 hours ago
$begingroup$
Thanks for the answer I believe I solved it based on first poster response as follows (in 1 kg). 700g P2O5 x 1 mol/ 142 g = 4.9 mol P2O5. 300 g H2O is 16.7 mol H2O. 4.9 mol P2O5 + 17.78 mol H2O --> 9.8 mol H3PO4 Mass of 9.8 mol H3PO4 * 98 g mol = 960.4 g H3PO4
$endgroup$
– dlight
7 hours ago
$begingroup$
Seems right, when dealing with percentage concentrations, it is better to use 100g as the unit. So using your numbers, it is 96% w/w phosphoric acid.
$endgroup$
– M. Farooq
7 hours ago
$begingroup$
Seems right, when dealing with percentage concentrations, it is better to use 100g as the unit. So using your numbers, it is 96% w/w phosphoric acid.
$endgroup$
– M. Farooq
7 hours ago
add a comment |
$begingroup$
$ceP2O5$ content is useful as it is the phosphorous content that a scientist often cares about when using phosphoric acids. This might seem silly given that as you pointed out anything from $0-70% ceP2O5$ could just be expressed as a percentage of $ceH3PO4$ but there are times when a higher $ceP2O5$ concentration is desired. For example polyphosphoric acids are useful reagents that have more phosphorous content that phosphoric acid. It would be ambiguous to state the concentration as $115% ceH3PO4$ (what does 115% mean) so instead procedures are reported in terms of $ceP2O5$ concentration with the balance assumed to be water (for those in the know).
answered 8 hours ago
A.K.A.K.
10.8k6 gold badges31 silver badges75 bronze badges
$endgroup$
add a comment |
$begingroup$
$ceP2O5$ content is useful as it is the phosphorous content that a scientist often cares about when using phosphoric acids. This might seem silly given that as you pointed out anything from $0-70% ceP2O5$ could just be expressed as a percentage of $ceH3PO4$ but there are times when a higher $ceP2O5$ concentration is desired. For example polyphosphoric acids are useful reagents that have more phosphorous content that phosphoric acid. It would be ambiguous to state the concentration as $115% ceH3PO4$ (what does 115% mean) so instead procedures are reported in terms of $ceP2O5$ concentration with the balance assumed to be water (for those in the know).
answered 8 hours ago
A.K.A.K.
10.8k6 gold badges31 silver badges75 bronze badges
$endgroup$
add a comment |
$begingroup$
$ceP2O5$ content is useful as it is the phosphorous content that a scientist often cares about when using phosphoric acids. This might seem silly given that as you pointed out anything from $0-70% ceP2O5$ could just be expressed as a percentage of $ceH3PO4$ but there are times when a higher $ceP2O5$ concentration is desired. For example polyphosphoric acids are useful reagents that have more phosphorous content that phosphoric acid. It would be ambiguous to state the concentration as $115% ceH3PO4$ (what does 115% mean) so instead procedures are reported in terms of $ceP2O5$ concentration with the balance assumed to be water (for those in the know).
answered 8 hours ago
A.K.A.K.
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$endgroup$
$ceP2O5$ content is useful as it is the phosphorous content that a scientist often cares about when using phosphoric acids. This might seem silly given that as you pointed out anything from $0-70% ceP2O5$ could just be expressed as a percentage of $ceH3PO4$ but there are times when a higher $ceP2O5$ concentration is desired. For example polyphosphoric acids are useful reagents that have more phosphorous content that phosphoric acid. It would be ambiguous to state the concentration as $115% ceH3PO4$ (what does 115% mean) so instead procedures are reported in terms of $ceP2O5$ concentration with the balance assumed to be water (for those in the know).
answered 8 hours ago
A.K.A.K.
10.8k6 gold badges31 silver badges75 bronze badges
answered 8 hours ago
A.K.A.K.
10.8k6 gold badges31 silver badges75 bronze badges
answered 8 hours ago
A.K.A.K.
10.8k6 gold badges31 silver badges75 bronze badges
answered 8 hours ago
A.K.A.K.
10.8k6 gold badges31 silver badges75 bronze badges
10.8k6 gold badges31 silver badges75 bronze badges
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1
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What's to explain? P2O5 reacts with water. Do you know what the product is? Can you balance the reaction?
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– Ivan Neretin
9 hours ago
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okay I didnt realize the concentration was in reference to water because its often discussed as purity etc. makes sense thanks.
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– dlight
9 hours ago
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@IvanNeretin Im reading a paper which states energy used to process phosphoric acid is "6500 kwh to make a ton of phosphoric acid (P2O5)". I want to express this in kWh/ton H3PO4. So I get 4710 kWh/ton H3PO4. Agreed? The paper is: A Technical Review of the Improved Hard Process
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– dlight
8 hours ago
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It is a common practice to express content of the component A in the equivalent content of the component B. There is an older unit of the water "hardness" dGH ( degree of General/German Hardness ) where 1 dGH means equivalent of 10 mg CaO / 1 L. But there is no CaO in water. Similarly, in fertilizers, there is AFAIK historical custom to use content of oxides, even if there are none.
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– Poutnik
8 hours ago
1
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P2O5 + 3H2O --> 2H3PO4. Using molecular weights P2O5/2H3PO4 = 142/198 = 0.71 = 71%
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– user55119
3 hours ago