help me balance this equation.....?

Question

P4 + HNO3 + H2O ----> H3PO4 + NO and its a redox reaction.

i dont need any physical or chemical changes....just how to balance it?
ive started but am stuck

just give me the final answer.!

Answer 1

There are several ways to balance redox reactions, and I prefer the half-reaction method.

Start with P4 on one side of an arrow, going to H3PO4 on the other. Since you have 4 atoms of P on the left, put a 4 in front of the phosphoric acid on the right. That gives you 16 oxygens on the right, so you need 16 water molecules on the left. That gives you 32 hydrogen atoms on the left, and you only have 12 on the right, so add 20 H+ on the right. Now, you have to balance the charge by adding 20 electrons to the right side.

Now, for the second half reaction, start with HNO3 on the left going to NO on the right. Your nitrogen is balanced. Next look at oxygen. You have 3 oxygen atoms on the left, and only one on the right so add 2 water molecules (H2O) on the right. That gives you 4 H atoms on the right, but you only have one on the left, so you need to add 3 H+ to the left side. Finally, balance the charges by adding 3 electrons to the left side.

Now, to get the correct over-all reaction, you need to make the number of electrons in the two half reactions the same. YOu can do that by multiplying everything in the top equation by 3 to give you size electrons, and everything in the second equation by 20 to again give you 60 electrons. Be careful when you do this to make sure you multiply ALL coefficients by the correct factor (3 or 20). Now, add the two equations together. You can simplify by subtracting anything that you have on both sides to get rid of it on one side or the other.

Answer 2

due to its imbalanced structure it would begin to bubble, and most likely change its physical and chemical properties

Answer 3

its very simple
3P4 + 20HNO3 + 8H2O = 12H3PO4 + 20NO

Answer 4

3P4 + 20HNO3 + 8H2O = 12H3PO4 + 20NO

Answer 5

try this website

http://www.webqc.org/balancedchemicalequations-061205-1.html