Very Tough Equation to Balance

[Visc]

NH₄ClO₄ + Al = HCl + Cl₂ + NO + Al₂O₃ + HOH

If anyone could give me some help and maybe try and explain it a little bit I would be very appreciative.

[expert]

First try to figure out how particular atoms reduce/oxidize. Make the following analysis
NH₄ClO₄ + Al = HCl + Cl₂ + NO + Al₂O₃ + HOH

2N(3-) - 10e- --> 2N(2+) – I doubled the nitrogen to make equal to total Cl – coming from the same molecule NH₄ClO₄
Cl(7+) + 7e- --> Cl(0)
Cl(7+) + 8e- --> Cl(1-)
Al(0) - 3e- --> Al(3+)

Basically, the complexity of this reaction is related to an unusual situation: it features two reducing agents and only one oxidizer. Still, the number of electrons given and accepted must stay equal. This provides a clue about basic coefficients. The total number of donated electrons is 10+3=13, and 7+8=15 accepted. The only coefficient can equalize them is 13x15=195. Nitrogen and Al must supply these electrons proportionally to their 10e- and 3e- loss, so let’s distribute these 195 electrons proportionally to demand by assigning multiplication coefficients:
2N(3-) - 10e- --> 2N(2+) x15 = 150e- = donated
Cl(7+) + 7e- --> Cl(0) x13=91e- - accepted
Cl(7+) + 8e- --> Cl(1-) x13=104e- accepted
Al(0) - 3e- --> Al(3+) x15=45e- - donated

Check: the number of donated and accepted electrons is equal and equal to the total of 195
The problem is that such coefficients will not provide equal numbers of N and Cl atoms – remember coming from the same molecule. We can only vary amount of Al.
Let’s equalize coefficients at N and Cl and let's absorb the difference by Al
2N(3-) - 10e- --> 2N(2+) x15 = 150e- = donated
Cl(7+) + 7e- --> Cl(0) x15=105e- - accepted
Cl(7+) + 8e- --> Cl(1-) x15=120e- accepted (total accepted 105+120=225; 225-150=75e- go to Al
Al(0) - 3e- --> Al(3+) x25=75e- - donated
Check again: the number of donated and accepted electrons is equal and equal to the total of 225
Now let’s put these found coefficients to play:
15x2NH₄ClO₄ + 25Al = 15HCl + 15/2Cl₂ + 30NO + 25/2Al₂O₃ + #HOH
Let’s multiply by 2 for convenience and calculate the number of protons to make water:
60NH₄ClO₄ + 50Al = 30HCl + 15Cl₂ + 60NO + 25Al₂O₃ + 105HOH

Finally oxygen check: 240 on the left and 240 on the right

Wow! That was the toughest one on this forum so far! Go collect your credit. I wonder where did you get this one?

By the way, this reaction has infinite number of solutions. That's why automatic program tells the following "Error: equation NH₄ClO₄+Al=HCl+Cl₂+NO+Al₂O₃+HOH can be balanced in an infinite number of ways: this is a combination of two different reactions"http://www.webqc.org/balance.php