8Fe + 13H2SO4 4Fe2(SO4)3 + H2S + 12H2O Cách cân bằng?

$\mathop{Fe}\limits^{0}+H_2\mathop{S}\limits^{+6}O_4 \to \mathop{Fe_2}\limits^{+3}(SO_4)_3+H_2\mathop{S}\limits^{-2}+H_2O$

$\text{ x 8 | }$$\mathop{Fe}\limits^{0} \to \mathop{Fe}\limits^{+3}+3e^-$

$\text{ x 3| }$$\mathop{S}\limits^{+6}+8e^- \to \mathop{S}\limits^{-2}$

`->`$8Fe+3\mathop{S}\limits^{+6} \to 8 \mathop{Fe}\limits^{+3}+3\mathop{S}\limits^{-2}$

`->`$8Fe+aH_2SO_4 \to 4Fe_2(SO_4)_3+3H_2S+bH_2O$

(Không lấy hệ số $\mathop{S}\limits^{+6}$ cho $H_2SO_4$ vì còn phụ thuộc vào $H$ nữa.)

Bảo toàn nguyên tố $S:$

$a=4 \times 3 +3$`=>`$a=15$

Bảo toàn nguyên tố $H:$

$2a=3 \times 2 +2b$`=>`$b=12$

`=>`$8Fe+15H_2SO_4 \to 4Fe_2(SO_4)_3+3H_2S+12H_2O$

`Fe + H_2SO_4 --> Fe_2(SO_4)_3 + H_2S + H_2O`

Oxi hóa - khử : 

`S^{+6} + 8e \to S^{-2} | × 3`

`Fe^0 \to Fe^{+3} + 3e | × 8 `

`⇒ 8Fe + 15H_2SO_4 \to 4Fe_2(SO_4)_3 + 3H_2S + 12H_2O`

$#Derix247$