Can you show me the steps to balancing this equation please?
HCl+HNF2=ClNF2+NH4Cl+HF
HCl+HNF2=ClNF2+NH4Cl+HF
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HCl + HNF2 → ClNF2 + NH4Cl + HF
Break it down into the two half-reactions:
HNF2 + 3H+ + 4e- → NH4+ + 2F-
(N^+ + 4e- → N^3-)
Cl- + HNF2 → ClNF2 + H+ + 2e-
(N^+ → N^3+ + 2e-)
Multiply the second one by 2 to balance the number of electrons with that of the first.
2Cl- + 2HNF2 → 2ClNF2 + 2H+ + 4e-
Now add the two together
2Cl- + 2HNF2 + HNF2 + 3H+ + 4e- → 2ClNF2 + 2H+ +4e- + NH4+ + 2F-
Cancel identical terms, and combine some (such as H+ + F- → HF):
2HCl + 3HNF2 + H+ → 2ClNF2 + NH4+ + 2HF
Now add Cl- to both sides and combine it with the H+ and NH4+:
3HCl + 3HNF2 → 2ClNF2 + NH4Cl + 2HF
Check the atom counts:
3Cl, 6H, 3N, 6F → 3Cl, 6H, 3N, 6F
Both sides have a net charge of zero, so the equaiton is balanced.
Break it down into the two half-reactions:
HNF2 + 3H+ + 4e- → NH4+ + 2F-
(N^+ + 4e- → N^3-)
Cl- + HNF2 → ClNF2 + H+ + 2e-
(N^+ → N^3+ + 2e-)
Multiply the second one by 2 to balance the number of electrons with that of the first.
2Cl- + 2HNF2 → 2ClNF2 + 2H+ + 4e-
Now add the two together
2Cl- + 2HNF2 + HNF2 + 3H+ + 4e- → 2ClNF2 + 2H+ +4e- + NH4+ + 2F-
Cancel identical terms, and combine some (such as H+ + F- → HF):
2HCl + 3HNF2 + H+ → 2ClNF2 + NH4+ + 2HF
Now add Cl- to both sides and combine it with the H+ and NH4+:
3HCl + 3HNF2 → 2ClNF2 + NH4Cl + 2HF
Check the atom counts:
3Cl, 6H, 3N, 6F → 3Cl, 6H, 3N, 6F
Both sides have a net charge of zero, so the equaiton is balanced.