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

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)

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

Sau phản ứng, thu được hỗn hợp kim loại, suy ra kẽm dư.

$n_{CuSO_4} = \dfrac{80.30\%}{160} = 0,15(mol)$

$Zn + CuSO_4 \to ZnSO_4 + Cu$

$n_{Zn\ pư} = n_{CuSO_4} = 0,15(mol)$
$\Rightarrow m_{Zn\ pư}  = 0,15.65 = 9,75(gam)$

Sau phản ứng, $m_{dd} = 9,75 + 80 - 0,15.64 = 80,15(gam)$

$C\%_{ZnSO_4} = \dfrac{0,15.161}{80,15}.100\% = 30,13\%$

n_CuSO4 = \(\dfrac{20.10\%}{160}\) = 0,0125 (mol)

Zn + CuSO₄ \(\rightarrow\) ZnSO₄ + Cu

0,0125 \(\leftarrow\) 0,0125 \(\rightarrow\) 0,0125 \(\rightarrow\) 0,0125 (mol)

m_{Zn pư} = 0,0125 . 65= 0,8125 (g)

m_{dd spư} = 20 + 0,8125 - 0,0125.64 = 20,0125 (g)

C%(ZnSO₄) = \(\dfrac{0,0125.161}{20,0125}\) . 100%= 10,06%

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

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)

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

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

Phản ứng trên là phản ứng thế

Ta có: \(n_{CuSO_4}=\dfrac{32\cdot10\%}{160}=0,02\left(mol\right)=n_{Cu}=n_{Zn}=n_{ZnSO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,02\cdot65=1,3\left(g\right)\\m_{Cu}=0,02\cdot64=1,28\left(g\right)\\m_{ZnSO_4}=0,02\cdot161=3,22\left(g\right)\\\end{matrix}\right.\) \(\Rightarrow C\%_{ZnSO_4}=\dfrac{3,22}{32+1,3-1,28}\cdot100\%\approx10,06\%\)

a)

$Zn + CuSO_4 \to ZnSO_4 + Cu$

b)

Theo PTHH : $n_{Zn} = n_{Cu} = n_{CuSO_4} = 0,25.2 = 0,5(mol)$
$m_{Zn} = 0,5.65 = 32,5(gam)$

$m_{Cu} = 0,5.64 = 32(gam)$

c)

Khối lượng thanh kẽm giảm là $32,5 - 32 = 0,5(gam)$

mCuSO4 = 4% x 200 = 8 (g)

=> nCuSO4 = 8/160 = 0,05 (mol)

PTHH : Zn + CuSO4 ------> ZnSO4 + Cu

a/ Từ PTHH ta có nZn = nCuSO4 = 0,05 (mol)

=> mZn = 0,05 x 65 = 3,25 (g)

b/ nZnSO4 = nCuSO4 = 0,05 (mol)

Suy ra mZnSO4 = 0,05 x 161 = 8,05 (g)

Vì Cu là chất kết tủa nên mdd = mZn + mddCuSO4 - mCu

= 3,25 + 200 - 0,05 x 64 = 200,05 (g)

=> C% = 8,05 / 200,05 x 100 = 4,024%

ncuso4=200*4/100/162=4/81 mol

zn+cuso4-> znso4+ cu

nzn=ncuso4=4/81

mzn=3,21g

mdd sau pư=200+3,21=203,21

C%znso4=161*100*4/81/203,21=3,91%