các bác làm hộ e bài này xem tn ạ

H

haisona8

T

thupham22011998

[TEX]2KMnO_4+16HCl-t*->2KCl+2MnCl_2+5Cl_2+8H_2O[/TEX]

[TEX]Cl_2+2NaOH-->NaCl+NaClO+H_2O[/TEX]

Ta có: n $KMnO_4$=0,02 mol --> n $HCl$=0,05 mol

n $NaOH$=0,2 mol

-->NaOH dư,$Cl_2$ pư hết

Theo pthh,ta có:n $NaCl$=n $NaClO$=0,05 mol

n $NaOH$ pư=0,1 mol -->n $NaOH$ dư=0,1 mol

-->CM $NaCl$=CM $NaClO$=0,05/0,2=0,25 M

-->CM $NaOH$ dư=0,1/0,2=0,5 M
 
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