Could you balance N2H4 plus O2-NO2 plus H2O?

Answer 1

How can balance NH3 plus O2 - NO plus H2O?

NH3 + O2 → NO + H2O

The easy way is to balance the most unbalanced atoms first.

3 H on left side, 2 H on right, make 6 H on both sides by multiplying 2 * NH3 and 3 * H2O

2 NH3 + O2 → NO +3 H2O

2 N on left, 1 N on right, make 2 N on right, 2 NO

2 NH3 + O2 → 2 NO +3 H2O

2 O on left, 5 O on right, make 10 O on both sides, 4 NO + 6 H2O

2 NH3 + 5 O2 → 4 NO +6 H2O

4 N on right, 2 N on left, make 4 N on left, 4 NO

4 NH3 + 5 O2 → 4 NO +6 H2O

Check

4N = 4N

12 H = 12 H

10 O = (4 + 6) 10 O

Ok

Now let me show a method that always works and is more organized.

Determine the charge of each atom. If necessary, check the electronegativity of an atom to determine if it is negative or positive.

For example in NH3, the N is on the left and the H is on the right, so you might think the N is positive and the H is negative. However the EN of H = 2.20 and the EN of N = 3.04. Since the EN of N is greater than the EN of H, the N has a stronger attraction for the pair of bonding electrons. So the N is considered negative and the H is positive.

H only has 2 electron, so it can only donate 1 electron to the N atom, so the charge on the H atom = +1, making the charge on N = -3.

N = -3

H = +1 ,as usual

N^-3(H^+1)3

When you write the charges, write them as superscripts!

The 3 at the end is a subscript.

Yahoo answers will not recognize and transmit superscripts of subscripts.

O2

O2 = (O^0)2

O2 = 0 charge

Any free (unbonded element) or one bonded to itself has a charge = 0.

NO

EN of O = 3.44

EN of N = 3.44

So O = -2

N = +2

N^+2 O^-2

H2O

O has greater EN, so

O = -2

H = +1

(H+1)2 O-2

Determine which element's atoms have changed charge as they changed from reactant to product.(left side to right side) To change charge, an atom either supplies electrons or attracts (gains) electrons

NH3 + O2 → NO + H2O

N^-3 → N^+2 in NO

To change from -3 to +2, the N atom gains 5 e-.

(O^0)2 → 2 O-2 in NO

For two O atoms to change from 0 to -2, each O atom must supply 2 e-, so O2 supplies 4 e-.

Here are the half-equations that show the electron exchange.

N^-3 → N^+2 + 5e-

(O^0)2 + 4 e- → 2 O-2

Notice N supplies 5 e- each time an N atom changes from -3 to +2.

Notice O gains 2 e- each time an O atom changes from 0 to -2, so O2 gains 4 e- each time an O2 changes from 0 to -2.

Like common denominators, for this exchange to occur one time. 20 e- most be exchanged. So, the N exchange occurs 4 times to supply enough e- for the O exchange to occur 5 times.

See the equations below:

4 N^-3 → 4 N^+2 + 20e-

5 (O^0)2 + 20 e- → 10 O-2

This 4 N's must appear in the original equation.

This 10 O's must appear in the original equation.

……….10 O……..4 O.+….6 O

…………..↓………...↓………↓

4 NH3 + 5 O2 → 4 NO +6 H2O

…↑……………...↑

4 N ………..….4N

Answer 2

The chemical equation is:
N2H4 + 3 O2 = 2 NO2 + 2 H2O