$\rm Zn+2AgNO_3 \xrightarrow{} Zn(NO_3)_2+2Ag$

Đặt $\rm n_{Zn}=a$ (mol) `=>` $\rm n_{Ag}=2a$ (mol)

Tăng giảm khối lượng:

$\rm 108 . 2a-65a=7.5$

`=>` $\rm a \approx 0.05$ (mol)

$\rm m_{Zn}=0.05 .65=3.25$ (g)

$\rm C_{M_{AgNO_3}}=\dfrac{0.05 . 2}{0.2}=0.5$ (M)

`#` NSD

`Zn + 2AgNO_3 -> Zn(NO_3)_2 + 2Ag`
Tăng giảm kl
`->` $n_{Zn}$ `=` $\dfrac{7,5}{2.108-65}$ `=0,05` mol
`->` $m_{Zn}$ `=0,05 . 65 = 3,25g`
$C_{MAgNO_3}$ `=` $\dfrac{0,05.2}{0,2}$ `=0,5M`