\(m_{FeCl_x}=\dfrac{32,5.10}{100}=3,25\left(g\right)\)
PTHH: FeClx + 3AgNO3 --> Fe(NO3)3 + xAgCl + (3-x)Ag
_______a---------------------------------------->ax----->a(3-x)
=> 143,5ax + 108a(3-x) = 8,61
=> 35,5ax + 324a = 8,61
=> a(35,5x+324) = 8,61
=> \(a=\dfrac{8,61}{35,5x+324}\)
=> \(M_{FeCl_x}=\dfrac{3,25}{\dfrac{8,61}{35,5x+324}}\)
=> 56 + 35,5x = \(\dfrac{325}{861}\left(35,5x+324\right)\)
=> x = 3
CTHH: FeCl3
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