cân bằng phản ứng oxi hóa - khử theo phương pháp thăng bằng electron 1H2SO4+H2S->S+H2O 2 s+HNO3->H2SO4+NO 3I2+HNO3->HIO3+NO+H2O 4 NH3+O2->No+H2O 5 C+HNO3->NO2+CO2+H2O 6H2SO4+HI->I2+H2S+H2O 7P+KClO3->P2O5+KCl 8 NH3+CuO->Cu+H2O+N2

Giải thích các bước giải:

1/.

$H_2S^{+6}O_4+H_2S^{-2}→S^0+H_2O$

$3$ x $|S^{-2}+2e→S^0$

$1$ x $|S^{+6}→S^0+6e$

$H_2SO_4+3H_2S→4S+4H_2O$

2/.

$S^0+HN^{+5}O_3→H_2S^{+6}O_4+N^{+2}O$

$1$ x $|S^0+6e→S^{+6}$

$2$ x $|N^{+5}→N^{+2}+3e$

$S+2HNO_3→H_2SO_4+2NO↑$

3/.

$I_2^0+HN^{+5}O_3→HI^{+5}O_3+N^{+2}O+H_2O$

$3$ x $|I_2^0+2.5e→I^{+5}$

$10$ x $|N^{+5}→N^{+2}+3e$

$3I_2+10HNO_3→6HIO_3+10NO↑+2H_2O$

4/.

$N^{-3}H_3+O_2^0→N^{+2}O^{-2}+H_2O$

$4$ x $|N^{-3}+5e→N^{+2}$

$5$ x $|O_2^0→2O^{-2}+2.2e$

$4NH_3+5O_2→4NO↑+6H_2O$

5/.

$C^0+HN^{+5}O_3→N^{+4}O_2+C^{+4}O_2+H_2O$

$1$ x $|C^0+4e→C^{+4}$

$4$ x $|N^{+5}→N^{+4}+1e$

$C+4HNO_3→4NO_2+CO_2↑+2H_2O$

6/.

$H_2S^{+6}O_4+HI^{-1}→I_2^0+H_2S^{-2}+H_2O$

$8$ x $|I^{-1}+1e→I_2^0$

$1$ x $|S^{+6}→S^{-2}+8e$

$H_2SO_4+8HI→4I_2+H_2S+4H_2O$

7/.

$P^0+KCl^{+5}O_3→P_2^{+5}O_5+KCl^{-1}$

$6$ x $|P^0+5e→P^{+5}$

$5$ x $|Cl^{+5}→Cl^{-1}+6e$

$6P+5KClO_3→3P_2O_5+5KCl$

8/.

$N^{-3}H_3+Cu^{+2}O→Cu^0+H_2O+N_2^0$

$2$ x $|N^{-3}+3e→N_2^0$

$3$ x $|Cu^{+2}→Cu^0+2e$

$2NH_3+3CuO→3Cu+3H_2O+N_2↑$