P4 + HNO3 = H3PO4 + NO2 + H20 balance by partial method
Indicate your specific subject in the "School Subject" box, so those with expertise in the area will respond to the question.
To balance the chemical equation P4 + HNO3 = H3PO4 + NO2 + H2O using the partial method, follow these steps:
Step 1: Split the equation
Separate the equation into its half-reactions. One representing the oxidation half-reaction (loss of electrons) and the other representing the reduction half-reaction (gain of electrons).
The given equation is: P4 + HNO3 = H3PO4 + NO2 + H2O
The oxidation half-reaction:
P4 → H3PO4
The reduction half-reaction:
HNO3 → NO2 + H2O
Step 2: Balance the atoms other than oxygen and hydrogen
Adjust the coefficients in each half-reaction to balance the atoms other than oxygen and hydrogen.
The oxidation half-reaction:
P4 → 4H3PO4
The reduction half-reaction:
HNO3 + 2H2O → NO2 + 4H2O
Step 3: Balance the oxygen atoms
Add water (H2O) molecules to the side that lacks oxygen atoms. The number of water molecules needed is determined by the difference in the number of oxygen atoms.
The oxidation half-reaction:
P4 → 4H3PO4 + 5H2O
The reduction half-reaction:
HNO3 + 2H2O → NO2 + 5H2O
Step 4: Balance the hydrogen atoms
Add hydrogen ions (H+) to the side that lacks hydrogen atoms. The number of hydrogen ions needed is determined by the difference in the number of hydrogen atoms.
The oxidation half-reaction:
P4 → 4H3PO4 + 5H2O
The reduction half-reaction:
HNO3 + 2H2O + 5H+ → NO2 + 5H2O
Step 5: Balance the charges
Add electrons (e-) to the side of the half-reaction that has a net positive charge. The number of electrons needed is determined by the difference in charge.
The oxidation half-reaction:
P4 + 20e- → 4H3PO4 + 5H2O
The reduction half-reaction:
HNO3 + 2H2O + 5H+ + 16e- → NO2 + 5H2O
Step 6: Combine the half-reactions
Multiply both half-reactions by appropriate coefficients so that the number of electrons are the same.
Multiplying the oxidation half-reaction by 16 and the reduction half-reaction by 5:
16P4 + 320e- → 64H3PO4 + 80H2O
5HNO3 + 10H2O + 25H+ + 80e- → 5NO2 + 25H2O
Step 7: Final balanced equation
Combine the balanced half-reactions to obtain the final balanced chemical equation.
16P4 + 5HNO3 + 18H2O → 64H3PO4 + 5NO2
P4 + HNO3 ---------> H3PO4 + NO +H2O
-> P4 + H2O -> H3PO4
Then using hit and trial method
P4 + H2O -> 4H3PO4 + 10[O]
(addition of 10 atoms of oxygen i.e. nascent oxygen to make the equation balanced)
-> HNO3 -> NO2 + H2O
Again using hit and trial method
2HNO3 +[O] -> 2NO2 + H2O
(addition of 1 atom of nascent oxygen to make the equation balanced)
NOTE : THE SAME EXTRA ATOM HAS TO BE MULTIPLIED ON BOTH THE SIDES AS PER THE SITUATION AND MUST BE CUT AT LAST
Now adding both the equations to make a single balanced equation
P4 + H2O -> 4H3PO4 + 10[O]
+ 2HNO3 +[O] -> 2NO2 + H2O } ×10
————————————————————
P4 + 20HNO3 -> 4H3PO4 + 2NO + 9H2O
P4 + HNO3 ---------> H3PO4 + NO +H2O
-> P4 + H2O -> H3PO4
Then using hit and trial method
P4 + H2O -> 4H3PO4 + 10[O]
(addition of 10 atoms of oxygen i.e. nascent oxygen to make the equation balanced)
-> HNO3 -> NO2 + H2O
Again using hit and trial method
2HNO3 +[O] -> 2NO2 + H2O
(addition of 1 atom of nascent oxygen to make the equation balanced)
NOTE : THE SAME EXTRA ATOM HAS TO BE MULTIPLIED ON BOTH THE SIDES AS PER THE SITUATION AND MUST BE CUT AT LAST
Now adding both the equations to make a single balanced equation
P4 + H2O -> 4H3PO4 + 10[O]
+ 2HNO3 +[O] -> 2NO2 + H2O } ×10
————————————————————
P4 + 20HNO3 -> 4H3PO4 + 2NO + 9H2O
AT last the oxygens and 1 water molecule get cancelled and the above answer remains
Hence done......=)