PTHH : \(Zn+CuSO_4\rightarrow ZnSO_4+Cu_{\downarrow}\)

$n_{AgNO_3} = \dfrac{150.6,8\%}{170} =0,06(mol)$
$Cu+ 2AgNO_3 \to Cu(NO_3)_2 + 2Ag$

Theo PTHH : 

$n_{Cu} = \dfrac{1}{2}n_{AgNO_3} = 0,03(mol)$
$m_{Cu} =0,03.64 = 1,92(gam)$

$n_{Ag} = n_{AgNO_3} = 0,06(mol)$

$\Rightarrow m_{dd\ sau\ pư} = 1,92 + 150 - 0,06.108 = 145,44(gam)$
$C\%_{Cu(NO_3)_2} = \dfrac{0,03.188}{145,44}.100\% = 3,88\%$

Ta có: \(m_{CuSO_4}=40.10\%=4\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{4}{160}=0,025\left(mol\right)\)

PT: \(Zn+CuSO_4\rightarrow ZnSO_4+Cu\)

Theo PT: \(n_{Zn}=n_{ZnSO_4}=n_{Cu}=n_{CuSO_4}=0,025\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,025.65=1,625\left(g\right)\)

Ta có: m _{dd sau pư} = 1,625 + 40 - 0,025.64 = 40,025 (g)

\(\Rightarrow C\%_{ZnSO_4}=\dfrac{0,025.161}{40,025}.100\%\approx10,056\%\)

Chọn D

Chọn D

\(n_{CuSO_4}=\dfrac{200.8}{100.160}=0,1(mol)\\ PTHH:Fe+CuSO_4\to FeSO_4+Cu\\ a,n_{Cu}=n_{Fe}=n_{CuSO_4}=0,1(mol)\\ \Rightarrow m_{Cu}=0,1.64=6,4(g);m_{Fe}=0,1.56=5,6(g)\\ b,n_{FeSO_4}=0,1(mol)\\ \Rightarrow C\%_{FeSO_4}=\dfrac{0,1.152}{5,6+200-6,4}.100\%=7,63\%\)

\(m_{ZnSO_4}=\dfrac{241,5.10}{100}=24,15\left(g\right)=>n_{ZnSO_4}=\dfrac{24,15}{161}=0,15\left(mol\right)\)

PTHH: 2Al + 3ZnSO₄ --> Al₂(SO₄)₃ + 3Zn

_____0,1<----0,15-------->0,05----->0,15

=> m_Al = 0,1.27 = 2,7(g)

=> m_Zn = 0,15.65=9,75(g)

b) mdd sau pư = 2,7 + 241,5 - 9,75 = 234,45(g)

=> \(C\%\left(Al_2\left(SO_4\right)_3\right)=\dfrac{0,05.342}{234,45}.100\%=7,294\%\)