Chemistry Forum

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.

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

Putting it all together, the decomposition products from ammonium perchlorate combine with aluminum, resulting in a highly exothermic reaction that produces HCl, Cl₂, NO, Al₂O₃, and H₂O.

The overall balanced reaction is complex, but for simplicity, you can think of it in terms of these main steps:

1. Ammonium perchlorate decomposes, generating reactive gases.
2. Aluminum burns in the presence of oxygen (from perchlorate), producing aluminum oxide and releasing energy.

This is why ammonium perchlorate and aluminum mixtures are often used in solid rocket propellants – the combination produces a high-temperature, high-energy reaction ideal for propulsion.

Don't answer 15 year old thread