a) \(n_{AgNO_3}=\dfrac{170.10\%}{170}=0,1\left(mol\right)\)
PTHH: Cu + 2AgNO3 ---> Cu(NO3)2 + 2Ag
0,05<--0,1--------->0,05--------->0,1
=> mCu (pư) = 0,05.64 = 3,2 (g)
b) mdd sau pư = 170 + 3,2 - 0,1.108 = 162,4 (g)
=> \(C\%_{Cu\left(NO_3\right)_2}=\dfrac{0,05.188}{162,4}.100\%=5,79\%\)
\(a,Cu+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2Ag\\ n_{AgNO_3}=\dfrac{170.10\%}{170}=0,1\left(mol\right)=n_{Ag}\\ n_{Cu}=n_{Cu\left(NO_3\right)_2}=n_{AgNO_3}:2=0,1:2=0,05\left(mol\right)\\ m_{Cu}=0,05.64=3,2\left(g\right)\\ b,m_{ddsau}=m_{Cu}+m_{ddAgNO_3}-m_{Ag}=3,2+170-0,1.108=162,4\left(g\right)\\ C\%_{ddCu\left(NO_3\right)_2}=\dfrac{188.0,05}{162,4}.100\approx5,788\%\)