Chất rắn chứa 3 kim loại nên Fe dư
\(Al+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3Ag\)
x/3_____ x_____________________
\(2Al+3Cu\left(NO_3\right)_2\rightarrow2Al\left(NO_3\right)_3+3Cu\)
\(\frac{0,03-x}{3}\rightarrow0,0045-0,5x\)
\(Fe+Cu\left(NO_3\right)_2\rightarrow Fe\left(NO_3\right)_2+Cu\)
1,5x-0,045__1,5x-0,045 __Fe còn dư = 0,095-1,5x
Ta có :
\(m_{kl}=x.108+\left(0,045-0,5x\right).64+\left(0,095-1,5x\right).56=8,12\)
\(\rightarrow x=0,01\)
\(\rightarrow CM=0,1M\)