### Balancing Chemical EquationsProblems #11 - 25

 Twenty examples Probs 1-10 Probs 26-45 Probs 46-65 "Balancing by groups" problems Only the problems quations Menu Balance redox equations by sight

Problem #11: NH3(g) + N2O(g) ---> N2(g) + H2O(g)

Solution:

1) Balance the H:

2NH3(g) + N2O(g) ---> N2(g) + 3H2O(g)

I picked hydrogen to start with because I knew I had to balance it with two coefficients at the same time. This is because H comes in threes on the left and in twos on the right. So, 2 x 3 = 3 x 2. I didn't pick oxygen to start with because it's already balanced. I didn't pick nitrogen because it's in three of the four compounds. Start with the easiest one first.

2) Balance the O:

2NH3(g) + 3N2O(g) ---> N2(g) + 3H2O(g)

3) Balance the N:

2NH3(g) + 3N2O(g) ---> 4N2(g) + 3H2O(g)

and it's done!

Problem #12: NH3 + O2 ---> NO + H2O

Solution:

1) Balance the H:

2NH3 + O2 ---> NO + 3H2O

2) Balance the N:

2NH3 + O2 ---> 2NO + 3H2O

3) Balance the O:

2NH3 + 52O2 ---> 2NO + 3H2O

4) Clear the fraction:

4NH3 + 5O2 ---> 4NO + 6H2O

Problem #13: NH3 + Cl2 ---> N2H4 + NH4Cl

Solution:

1) Chlorine is the simplest one to balance, so do it first:

NH3 + Cl2 ---> N2H4 + 2NH4Cl

2) Balance the N:

4NH3 + Cl2 ---> N2H4 + 2NH4Cl

This also balances the hydrogen. We're done!

Suppose you had picked the nitrogen first. You might have tried a 3 in front of the N on the left, but then seen that that gave an odd number of H when you know that the number of H on the right must always be even. Maybe you then bump up the coefficient to 4 and then . . . .

Of course, this is why the chlorine is simpler. It's in one place on each side whereas the N is in one place on the left, but two places on the right. Start with chlorine!

Problem #14: NH3 + F2 ---> N2F4 + HF

Solution:

1) Balance N:

2NH3 + F2 ---> N2F4 + HF

2) Balance H:

2NH3 + F2 ---> N2F4 + 6HF

3) Balance F:

2NH3 + 5F2 ---> N2F4 + 6HF

Reminder: always try to balance the simplest element first. By simplest, I mean the element that appears in only one compound on each side. You could pick H in the above example and, if you did, you'd go back and fix it after then balancing the nitrogens. A person with lots of experience balancing would usually start with nitrogen (based just on experience), but starting with hydrogen only makes the sequence one step longer.

Problem #15: NH3 + F2 ---> NH4F + NF3

Solution:

1) Pick the hydrogen first because it's the simplest one to balance (being in only one compound on each side:

4NH3 + F2 ---> 3NH4F + NF3

2) This also balances the nitrogen. Finish with the fluorine:

4NH3 + 3F2 ---> 3NH4F + NF3

Problem #16: NH4NO3(s) ---> N2(g) + O2(g) + H2O(g)

Solution:

1) Balance H:

NH4NO3(s) ---> N2(g) + O2(g) + 2H2O(g)

2) Balance O:

NH4NO3(s) ---> N2(g) + 12O2(g) + 2H2O(g)

3) Multiply through to clear fraction:

2NH4NO3(s) ---> 2N2(g) + O2(g) + 4H2O(g)

Note: nitrogen was always in balance, so it was never addressed.

Problem #17: N2O + CH4 ---> N2 + CO2 + H2O

Comment: the H and the O are interconnected because when you go to balance the H, that will affect the O.

Solution:

1) Balance the H first:

N2O + CH4 ---> N2 + CO2 + 2H2O

2) Now, balance the O:

4N2O + CH4 ---> N2 + CO2 + 2H2O

3) And finish with the N:

N2O + CH4 ---> 4N2 + CO2 + 2H2O

Problem #18: NO2 + O2 + H2O ---> HNO3

Solution #1:

1) Balance H:

NO2 + O2 + H2O ---> 2HNO3

2) Balance N:

2NO2 + O2 + H2O ---> 2HNO3

3) Balance O (by taking away an O with a fraction):

2NO2 + 12O2 + H2O ---> 2HNO3

4) Multiply through by 2:

4NO2 + O2 + 2H2O ---> 4HNO3

Solution #2:

1) Balance H:

NO2 + O2 + H2O ---> 2HNO3

2) Notice that, with a 2 in front of HNO3, the total O on the left MUST be even because the coefficient in front of HNO3 MUST be even, so as to balance H (since the H on the left only comes in even amounts). Do this:

NO2 + O2 + H2O ---> 4HNO3

3) The reason is because the contribution of O from H2O must be an even amount, consequently we must have an even coefficient in front of the H2O. Balance H:

NO2 + O2 + 2H2O ---> 4HNO3

4) Balance N:

4NO2 + O2 + 2H2O ---> 4HNO3

Problem #19: NO2 + H2O ----> HNO3 + NO

Solution #1:

1) The H is in one compound on the left and one compound on the right. Balance it first:

NO2 + H2O ---> 2HNO3 + NO

2) Balance the N:

3NO2 + H2O ---> 2HNO3 + NO

3) As a result of balacing the nitrogen, the oxygens also are balanced. We are done!

Solution #2:

1) Balance N:

2NO2 + H2O ----> HNO3 + NO

2) Balance H:

2NO2 + H2O ----> 2HNO3 + NO

3) Rebalance N:

3NO2 + H2O ----> 2HNO3 + NO

4) Check oxygens: seven on each side. Oxygen is balanced. Done!

Problem #20: N2(g) + O2(g) + H2O ---> HNO3

Solution:

1) Balance the N:

N2(g) + O2(g) + H2O --> 2HNO3

2) Notice how that balances the H, so do the O:

N2(g) + 52O2(g) + H2O --> 2HNO3

3) That 52O2 is equal to 5 oxygen atoms, so add that to the one O in the H2O and the O is balanced. However, we are not done:

2N2(g) + 5O2(g) + 2H2O --> 4HNO3

Now we're done!

Problem #21: N2H4 + N2O4 ---> N2 + H2O

Solution #1:

1) Balance the O:

N2H4 + N2O4 ---> N2 + 4H2O

2) Balance the H:

2N2H4 + N2O4 ---> N2 + 4H2O

3) Balance the N:

2N2H4 + N2O4 ---> 3N2 + 4H2O

Solution #2:

1) Since all of the hydrogen and all of the oxygen go to make H2O, we will need double the H on the left-hand side compared to oxygen on the left-hand side:

2N2H4 + N2O4 ---> N2 + H2O

2) Balance the H and the O:

2N2H4 + N2O4 ---> N2 + 4H2O

3) Balance N:

2N2H4 + N2O4 ---> 3N2 + 4H2O

Problem #22: N2H4 + H2O2 ---> N2 + H2O

Solution #1:

1) The nitrogen is balanced. The oxygen is in one place on each side. Balance it:

N2H4 + H2O2 ---> N2 + 2H2O

2) The hydrogens are not balanced. And notice that the oxygens on the left side only come in twos. So, do this:

N2H4 + H2O2 ---> N2 + 4H2O

3) Fix the oxygen and see what happens to the hydrogen:

N2H4 + 2H2O2 ---> N2 + 4H2O

Eight hydrogens on each side. The equation is balanced.

Solution #2:

1) Balance the oxygen, as above:

N2H4 + H2O2 ---> N2 + 2H2O

2) Balance the hydrogens with a fractional coefficient:

12N2H4 + H2O2 ---> N2 + 2H2O

3) That messes up the balance of nitrogen, so fix it:

12N2H4 + H2O2 ---> 12N2 + 2H2O

4) Multiply through to clear the fraction.

Problem #23: N2O4 + H2O ---> HNO3 + NO

Solution:

1) The nitrogens are balanced, but the hydrogens are not. Balance the hydrogens (and ignore the oxygen for the moment):

N2O4 + H2O ---> 2HNO3 + NO

2) This throws the nitrogen out of balance, so fix it (still ignoring the oxygens):

32N2O4 + H2O ---> 2HNO3 + NO

3) Nitrogen and hydrogen both balanced. Check the oxygens to find 7 on each side, so it's balanced as well. Clear the fraction:

3N2O4 + 2H2O ---> 4HNO3 + 2NO

Problem #24: NaOH + Al --> Al2O3 + H2 + Na2O

Solution:

1) Balance the Al:

NaOH + 2Al --> Al2O3 + H2 + Na2O

2) Balance the Na:

2NaOH + 2Al --> Al2O3 + H2 + Na2O

3) This balances the hydrogen, but the oxygen is still not balanced. Since oxygen is only in the NaOH on the left, try this:

4NaOH + 2Al --> Al2O3 + H2 + 2Na2O

4) Oxygen is still not balanced, but notice that, in the first try, the oxygen difference was two too many on the right. In the second try, the difference is one too many on the right. So, try this:

6NaOH + 2Al --> Al2O3 + H2 + 3Na2O

5) The oxygen is balanced. Finish with the hydrogen:

6NaOH + 2Al --> Al2O3 + 3H2 + 3Na2O

This situation where the difference in amounts getting smaller (seen in step 4) occurs from time to time. Be aware of it.

Problem #25: KMnO4 ---> K2O + MnO + O2

Solution:

1) Balance K:

2KMnO4 ---> K2O + MnO + O2

2) Balance Mn:

2KMnO4 ---> K2O + 2MnO + O2

3) Balance O:

2KMnO4 ---> K2O + 2MnO + 52O2

Eight O on the left and, on the right, one in the K2O and two in the 2MnO, so five more needed. 52O2 is 5 oxygen atoms.

4) Multiply through by 2 to clear the fraction:

4KMnO4 ---> 2K2O + 4MnO + 5O2

Bonus Problem: CaF2 3Ca3(PO4)2 + H2SO4 + H2O ---> H3PO4 + HF + CaSO4 2H2O

Solution:

1) Balance the calcium:

CaF2 3Ca3(PO4)2 + H2SO4 + H2O ---> H3PO4 + HF + 10CaSO4 2H2O

2) Balance the phosphorous:

CaF2 3Ca3(PO4)2 + H2SO4 + H2O ---> 6H3PO4 + HF + 10CaSO4 2H2O

3) Balance the sulfur:

CaF2 3Ca3(PO4)2 + 10H2SO4 + H2O ---> 6H3PO4 + HF + 10CaSO4 2H2O

4) Balance the fluorine:

CaF2 3Ca3(PO4)2 + 10H2SO4 + H2O ---> 6H3PO4 + 2HF + 10CaSO4 2H2O

5) Balance the hydrogens. Count them on the right-hand side to find 60 (18 in the phosphoric acid, 2 in the HF and 40 in the calcium sulfate dihydrate). On the left-hand side, there are already 20 from the sulfuric acid, so we need 40 hydrogens from the water:

CaF2 3Ca3(PO4)2 + 10H2SO4 + 20H2O ---> 6H3PO4 + 2HF + 10CaSO4 2H2O

There are 84 oxygens on each side. It's balanced.

 Twenty examples Probs 1-10 Probs 26-45 Probs 46-65 "Balancing by groups" problems Only the problems Equations Menu Balance redox equations by sight