P4 + HNO3 = H3PO4 + NO2 + H20 by partial method balancing

Question

Create a visually stimulating depiction of a chemistry experiment. Display a flask filled with a vibrant P4 substance on the left. Next to it, place a bottle of HNO3 with a vivid liquid. Have these substances interacting in the center, forming a beautiful reaction. On the right, show the resultant compounds, H3PO4, NO2, and H2O, in separate containers. Make sure to convey the process of partial method balancing, possibly using suggestive shapes or colors. However, the image should contain absolutely no text.

P4 + HNO3 = H3PO4 + NO2 + H20 by partial method balancing

Answer 1

This answer is wrong

Answer 2

Why did the P4 and HNO3 go to couples therapy?

Because they couldn't find a balanced equation on their own!

Answer 3

So, first step: decide on the partial equations.

There's a good discussion here:

http://chem-guide.blogspot.com/2010/03/balancing-chemical-equations.html

Answer 4

To balance the chemical equation P4 + HNO3 = H3PO4 + NO2 + H2O using the partial method, you need to balance the atoms in the equation one at a time. Here's how you can do it:

Step 1: Start by balancing the most complex or least abundant atoms first. In this case, we will start with phosphorus (P) and nitrogen (N).

Count the number of phosphorus atoms on both sides of the equation:
Left Side: 4 P
Right Side: 1 P (in H3PO4)

To balance the phosphorus, we need to place a coefficient of 4 in front of H3PO4:
P4 + HNO3 = 4 H3PO4 + NO2 + H2O

Now, let's balance the nitrogen (N) atoms:

Count the number of nitrogen atoms on both sides of the equation:
Left Side: 1 N (in HNO3)
Right Side: 1 N (in NO2)

To balance the nitrogen, we need to place a coefficient of 2 in front of NO2:
P4 + HNO3 = 4 H3PO4 + 2 NO2 + H2O

Next, let's balance the hydrogen (H) and oxygen (O) atoms:

Count the number of hydrogen atoms on both sides of equation:
Left Side: 1 H (in HNO3)
Right Side: 12 H (in H3PO4) + 2 H (in H2O)

To balance the hydrogen, we need to place a coefficient of 12 in front of H3PO4 and a coefficient of 2 in front of H2O:
P4 + HNO3 = 4 H3PO4 + 2 NO2 + 12 H2O

Now, let's balance the oxygen:

Count the number of oxygen atoms on both sides of the equation:
Left Side: 3 O (in HNO3)
Right Side: 12 O (in H3PO4) + 2 O (in NO2) + 12 O (in H2O)

To balance the oxygen, we need to place a coefficient of 6 in front of HNO3:
P4 + 6 HNO3 = 4 H3PO4 + 2 NO2 + 12 H2O

Finally, check if all the atoms are balanced. In this case, there are 4 phosphorus (P), 6 nitrogen (N), 24 hydrogen (H), and 24 oxygen (O) atoms on both sides.

Thus, the balanced equation is:
P4 + 6 HNO3 = 4 H3PO4 + 2 NO2 + 12 H2O

Answer 5

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......=)