[Hóa 10]Cân bằng phương trình phản ứng

E

endinovodich12

Chúc noen vui vẻ !

a;
- Quá trình oxi hóa : 1 | [tex]2C^-2 -----------> 2C^+2 + 8 e [/tex]
- Quá trình khử : 4 | [tex] I^0 + 2e -------> 2I^-1 [/tex]

Ta cân bằng như sau:

[tex] C_2H_5OH +6 NaOH + 4I_2 --------> CHI_3 + HCOONa + 5NaI +5H_2O [/tex]
 
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E

endinovodich12

Cần gấp!

Ta có :

*********| [tex]2x C^-(\frac{y-2}{x}[/tex] ---------->[tex] 2x C^-1 + 2(y-2+x)e [/tex]
*********| [tex]x C^-(\frac{y-2}{x})[/tex] ---------->[tex] xC^+4 + (y-2-4x)e [/tex]
--------------------------------------------------------------------------------------------------------------
*********5|[tex]3x C^-(\frac{y-2}{x})[/tex] ---------->[tex]2xC^-1 + xC^+4 +(3y-6-2x)e[/tex]
[tex] (3y-6-2x)|Mn^+7 +5e --------------> Mn^+2 [/tex]

[tex]15C_xH_yO + (3y-6-2x)KMnO_4 + (9y-18-6x)HCl [/tex] --------------------------------->
[tex]5xC_2H_4O +(3y-6-2x)MnCl_2 +5xCO_2 + (3y-6-2x)KCl + (12y-9-13x)H_2O [/tex]
 
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