\(n_{H_2SO_4}=0,6\left(mol\right)\)

\(m_{BaCl_2}=80\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{5}{13}\left(mol\right)\)

\(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)

Dễ thấy \(H_2SO_4\) dư 

\(\Rightarrow n_{BaSO_4\downarrow}=n_{BaCl_2}=\dfrac{5}{13}\left(mol\right)\)

Mà hiệu suất chỉ đạt 80%

\(\Rightarrow n_{BaSO_4\downarrow}=\dfrac{5}{13}.\dfrac{80}{100}=\dfrac{4}{13}\left(mol\right)\)

\(\Rightarrow m_{BaSO_4\downarrow}=71,69\left(g\right)\)

mZn= 3,25/65=0,05(mol)

mddCuSO4=1,12.40=44,8(g)

-> mCuSO4= 44,8. 25%=11,2(g) => nCuSO4= 11,2/160=0,07(mol)

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

Ta có: 0,05/1 < 0,07/1

=> Zn hết, CuSO4 dư, tính theo nZn.

=> nCuSO4(p.ứ)=nCu=nZnSO4=nZn=0,05(mol)

=>nCuSO4(dư)=0,07-0,05=0,02(mol)

Vddsau= VddCuSO4=0,04(l)

=> CMddCuSO4(dư)= 0,02/0,04=0,5(M)

CMddZnSO4=0,05/0,04=1,25(M)

c) m(muối)= mCuSO4(dư)+ mZnSO4= 0,02. 160+ 0,05.161=11,25(g)

a)
\(Fe_2\left(SO_4\right)_3+Zn\rightarrow2FeSO_4+ZnSO_4\)

\(CuSO_4+Zn\rightarrow ZnSO_4+Cu\downarrow\)

\(FeSO_4+Zn\rightarrow ZnSO_4+Fe\downarrow\)

b)
\(m_O=\dfrac{83,6.45,934}{100}=38,4\left(g\right)\)

=> \(n_O=\dfrac{38,4}{16}=2,4\left(mol\right)\)

=> \(n_{SO_4}=\dfrac{2,4}{4}=0,6\left(mol\right)\)

=> \(n_{ZnSO_4}=0,6\left(mol\right)\)

=> \(n_{Zn\left(pư\right)}=0,6\left(mol\right)\)

\(m_{Fe,Cu}=83,6-0,6.96=26\left(g\right)\)

m_{thanh KL (sau pư)} = 100 + 26 - 0,6.65 = 87(g)

theo tui là 0,15 M

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,\(n_{H_2SO_4}=0,2.3=0,6\left(mol\right);n_{BaCl_2}=0,35.2=0,7\left(mol\right)\)

PTHH: H₂SO₄ + BaCl₂ → BaSO₄ ↓ + 2HCl

Mol:       0,6          0,6           0,6            0,12

Ta có: \(\dfrac{0,6}{1}< \dfrac{0,7}{1}\)⇒ H₂SO₄ hết, BaCl₂ dư

\(m_{BaSO_4}=0,6.233=139,8\left(g\right)\)

b,V_{dd sau pứ} = 0,2+0,35 = 0,55 (l)

\(C_{M_{HCl}}=\dfrac{0,6}{0,55}=\dfrac{12}{11}M\)

\(C_{M_{BaCl_2dư}}=\dfrac{0,7-0,6}{0,55}=\dfrac{2}{11}M\)

c,\(m_{H_2SO_4\left(lt\right)}=0,6.98=58,8\left(g\right)\Rightarrow m_{H_2SO_4\left(pứ\right)}=\dfrac{58,8}{75\%}=78,4\left(g\right)\)

\(n_{H_2SO_4}=\dfrac{78,4}{98}=0,8\left(mol\right)\)

PTHH: 3FeS₂ + 6H₂O + 11O₂ → Fe₃O₄ + 6H₂SO₄

Mol:      0,4                                                       0,8

\(m_{FeS_2\left(lt\right)}=0,8.120=96\left(g\right)\)

0từ đề bài bạn tính được số mol CuSO4 là 0.2 x 0.1 = 0.02 mol

PTHH: Zn + CuSO4 -> Cu + ZnSO4

Từ PTTT số mol CuSO4 = số mol Zn = 0.02

Khối lượng kẽm phản ứng là: 0,02 x 65 = 1.3 g

\(n_{CuSO_4}=C.V=0,1.0,2=0,02\left(mol\right)\)
PTHH: \(Zn+CuSO_4\rightarrow ZnSO_4+Cu\)
Theo PT ta có: \(n_{Zn}=n_{CuSO_4}=0,02\left(mol\right)\)
\(\Rightarrow m_{Zn\left(pư\right)}=0,02.65=1,3\left(g\right)\)