[QUOTE=dangtrinhlvl;63768]
giai bang DTBL e
3/ hoa tan hh (0,1mol Ag va 0,04mol Cu) vao dd HNO3 thu dc hh khi X (NO va NO2) ti le so mol tuong ung la 2 : 3. tinh V hh khi X? (em ra 2,24l ko biet dung ko)
ta co so do: 5N(+5) + 11e-----> 2N(+4) + 3N(+2)
n echo = n Ag + 2.n Cu = 0,18 (mol)
số mol e để N(+5) ----> N(+4) la: 2.0,18/11 = 9/275 (mol)
=> n NO2 = 9/275.1 = 9/275 (mol)
số mol e dể N(+5) ----> N(+4) la : 0,18 - 9/275 = 81/550 ( mol)
=> n NO = 81/(550.2)= 81/1100 mol
V = (9/275 + 81/1100) . 22,4 = 2,383 l
co van de j thi liên lac wa nick
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