\(n_{Cu}=\dfrac{m}{M}=\dfrac{32}{64}=0,5mol\)
\(n_{AgNO_3}=0,5mol\)
Cu+2AgNO3\(\rightarrow\)Cu(NO3)2+2Ag
x\(\rightarrow\)2x.....................x............2x
-Độ tăng khối lượng=108.2x-64x=62,4-32=30,4
\(\rightarrow\)152x=30,4\(\rightarrow\)x=0,2mol
-Dung dịch Y có:
+ \(n_{Cu\left(NO_3\right)_2}=x=0,2mol\)\(\rightarrow C_M=\dfrac{0,2}{0,5}=0,4M\)
+\(n_{AgNO_3\left(dư\right)}=0,5-2x=0,5-0,4=0,1mol\)
\(\rightarrow C_M=\dfrac{0,1}{0,5}=0,2M\)