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# A quicker way to balance an equation? watch

1. Can anyone tell me a quick way to balance this equation. Some people say that you should start with the non mental first and leave H and O at last but any suggestions?? What is the quickest way to balance an equation like this: P4+H2O ---> PH3 +H3PO4 than you.
2. Leave the elements til last if there are any (like the P4 in your equation above).
Do the O's first because they only appear in one compound on each side so you can just adjust the ratios. Then fix the H ratios by changing the PH3. Since you've left the element til last it'll now be easy to change the P4 to whatever it needs to be. I also find it's often easier also to just leave things in fractions and then multiply out the fractions at the end:

P4+H2O ---> PH3 +H3PO4 ----------------- Unbalanced equation
P4+4H2O ---> PH3 +H3PO4 --------------- change the O ratios (1 to 4)
P4+4H2O ---> (5/3)PH3 +H3PO4 -------- 8 H on LHS so there should be 8 Hs on the RHS
(2/3)P4+4H2O ---> (5/3)PH3 +H3PO4 --- 8/3 Ps on RHS, so should be 8/3 Ps on LHS

Both sides are now equal, so multiply both sides through by 3 to remove fractions:

2P4+12H2O ---> 5PH3 +3H3PO4

and done
3. (Original post by coconut64)
Can anyone tell me a quick way to balance this equation. Some people say that you should start with the non mental first and leave H and O at last but any suggestions?? What is the quickest way to balance an equation like this: P4+H2O ---> PH3 +H3PO4 than you.
This is a redox reaction (actually disproportionation)

Write down each half equation using oxidation numbers to decide how many electrons are transferred.

equation 1 ===> P + 3H+ +3e --> PH3
equation 2 ===> P + 4H2O --> H3PO4 + 5H+ + 5e

now equalise electrons

from equation 1 ===> 5P + 15H+ + 15e --> 5PH3
from equation 2 ===> 3P + 12H2O --> 3H3PO4 + 15H+ + 15e
8P + 12H2O --> 5PH3 + 3H3PO4

put the phosphorus into molecules

2P4 + 12H2O --> 5PH3 + 3H3PO4
4. (Original post by charco)
This is a redox reaction (actually disproportionation)

Write down each half equation using oxidation numbers to decide how many electrons are transferred.

equation 1 ===> P + 3H+ +3e --> PH3
equation 2 ===> P + 4H2O --> H3PO4 + 5H+ + 5e

now equalise electrons

from equation 1 ===> 5P + 15H+ + 15e --> 5PH3
from equation 2 ===> 3P + 12H2O --> 3H3PO4 + 15H+ + 15e
8P + 12H2O --> 5PH3 + 3H3PO4

put the phosphorus into molecules

2P4 + 12H2O --> 5PH3 + 3H3PO4
I haven't learnt how to balance the equation using oxidation number yet but is there a simple way to work this out? Thanks
5. (Original post by Pronged Lily)
Leave the elements til last if there are any (like the P4 in your equation above).
Do the O's first because they only appear in one compound on each side so you can just adjust the ratios. Then fix the H ratios by changing the PH3. Since you've left the element til last it'll now be easy to change the P4 to whatever it needs to be. I also find it's often easier also to just leave things in fractions and then multiply out the fractions at the end:

P4+H2O ---> PH3 +H3PO4 ----------------- Unbalanced equation
P4+4H2O ---> PH3 +H3PO4 --------------- change the O ratios (1 to 4)
P4+4H2O ---> (5/3)PH3 +H3PO4 -------- 8 H on LHS so there should be 8 Hs on the RHS
(2/3)P4+4H2O ---> (5/3)PH3 +H3PO4 --- 8/3 Ps on RHS, so should be 8/3 Ps on LHS

Both sides are now equal, so multiply both sides through by 3 to remove fractions:

2P4+12H2O ---> 5PH3 +3H3PO4

and done
So I should always start with element that only comes up once on both sides? Also, i never tried using fractions to balance an equation, could you give a summary of how to balance an equation (any equation, not just this one). Thanks for the help, this is actually really helpful!
6. (Original post by coconut64)
I haven't learnt how to balance the equation using oxidation number yet but is there a simple way to work this out? Thanks
no, this is the correct method and can always be used for reactions involving redox regardless of complication.

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