\(n_{CuO}=\dfrac{16}{80}=0.2\left(mol\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(0.2........0.2..................0.2\)

\(V_{dd_{H_2SO_4}}=\dfrac{0.2}{1}=0.2\left(l\right)\)

\(m_{CuSO_4}=0.2\cdot160=32\left(g\right)\)

\(n_{Fe}=\dfrac{22,4}{56}=0,4\) (mol) (1)

Phương trình hóa học : 

Fe + 2HCl ---> FeCl₂ + H₂ (2) 

Từ (1) và (2) ta có \(n_{FeCl_2}=n_{H_2}=0,4\) (mol) ; \(n_{HCl}=0,8\left(mol\right)\)

b) => \(m_{\text{muối}}=0,4.\left(56+35,5.2\right)=50.8\left(g\right)\)

c) \(V_{\text{khí}}=0,4.22,4=8,96\left(l\right)\)

d) \(m_{HCl}=0,8.36.5=29,2\left(g\right)\)

\(\Rightarrow C\%=\dfrac{29,2}{200}.100\%=14,6\%\)

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(0.2.........0.2.............0.2\)

\(m_{ZnSO_4}=0.2\cdot161=32.2\left(g\right)\)

\(V_{dd_{H_2SO_4}}=\dfrac{0.2}{0.5}=0.4\left(l\right)\)

\(a) Zn + H_2SO_4 \to ZnSO_4 + H_2\\ n_{ZnSO_4} = n_{H_2SO_4} = n_{Zn} = \dfrac{13}{65} = 0,2(mol)\\ m_{ZnSO_4} = 0,2.161 = 32,2(gam)\\ b) V_{dd\ H_2SO_4}= \dfrac{0,2}{0,5} = 0,4(lít)\)

a. Ta có nZn= \(\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn+H2SO4--->ZnSO4+H2

  => nZn=nZnSO4=0.2mol

=> mZnSO4=0,2* 161=32.2(g)

b. Theo pthh trên, nZn=nH2SO4=0.2 mol

=> VH2SO4= 0.2:0.5= 0,4(lít)

Bài 2 : 

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

PTHH :

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

0,1           0,3                 0,1             0,3

\(m_{Fe_2\left(SO_4\right)_3}=0,1.400=40\left(g\right)\)

\(b,V_{ddH_2SO_4}=\dfrac{0,3}{2}=0,15\left(l\right)\)

\(c,C_{M\left(Fe_2\left(SO_4\right)_3\right)}=\dfrac{0,1}{0,15}=\dfrac{2}{3}\left(M\right)\)

Bài 3 :

\(n_{Mg}=\dfrac{4.8}{24}=0,2\left(mol\right)\)

PTHH :

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2O\)

0,2       0,2              0,2          0,2

\(m_{MgSO_4}=0,2.120=24\left(g\right)\)

\(V_{H_2}=0,2.24,79=4,958\left(l\right)\)

\(c,C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,2}=1\left(M\right)\)

\(d,C_{M\left(MgSO_4\right)}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Bài 4 :

\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

PTHH :

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

0,3       0,45              0,15          0,45

\(V_{H_2}=0,45.24,79=11,1555\left(l\right)\)

\(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\)

\(C\%_{H_2SO_4}=\dfrac{44,1}{300}.100\%=14,7\%\)

\(m_{Al_2\left(SO_4\right)_3}=0,15.342=51,3\left(g\right)\)

\(m_{dd}=8,1+300-\left(0,45.2\right)=307,2\left(g\right)\)

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{51,3}{307,2}.100\%\approx16,7\%\)

Bài 5 :

\(n_{H_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

PTHH:

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

0,2        0,4        0,2       0,2

\(m_{Fe}=0,2.56=11,2\left(g\right)\)

\(C_{M\left(HCl\right)}=\dfrac{0,4}{0,2}=2\left(M\right)\)

\(C_{M\left(FeCl_2\right)}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Bài 6 : 

\(n_{Fe\left(OH\right)_2}=\dfrac{1,8}{90}=0,02\left(mol\right)\)

PTHH :

\(Fe\left(OH\right)_2+2HCl\rightarrow FeCl_2+2H_2O\)

0,02             0,04       0,02          0,04

\(m_{HCl}=0,04.36,5=1,46\left(g\right)\)

\(C\%_{HCl}=\dfrac{1,46}{200}.100\%=0,73\%\)

\(m_{FeCl_2}=0,02.127=2,54\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{2,54}{1,8+200}\approx1,259\%\)

Bài 7 : 

\(n_{NaOH}=0,1.2=0,2\left(mol\right)\)

PTHH :

\(NaOH+HCl\rightarrow NaCl+H_2O\)

0,2           0,2          0,2          0,2

\(a,m_{NaCl}=0,2.58,5=11,7\left(g\right)\)

\(b,C_{M\left(HCl\right)}=\dfrac{0,2}{0,1}=2\left(M\right)\)

\(c,C_{M\left(NaCl\right)}=\dfrac{0,2}{0,1+0,1}=1\left(M\right)\)

a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)

c, \(n_{H_2SO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{FeSO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow m_{FeSO_4}=0,35.152=53,2\left(g\right)\)

e, \(C_{M_{FeSO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{H_2SO_4}=0,25.1,6=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)

a,\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: Fe + H₂SO₄ → FeSO₄ + H₂

Mol:     0,1       0,1          0,1        0,1

b,\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c,\(C_{M_{ddH_2SO_4}}=\dfrac{0,1}{0,2}=0,5M\)

d,\(C_{M_{ddFeSO_4}}=\dfrac{0,1}{0,2}=0,5M\)

ai giúp mình với

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ a.Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ b.0,1.......0,1..........0,1...........0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c.C_{MddH_2SO_4}=\dfrac{0,1}{0,2}=0,5\left(M\right)\\ d.C_{MddFeSO_4}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

Sửa đề : 11.2 g sắt

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(0.2....0.2.................0.2\)

\(m_{FeSO_4}=0.2\cdot152=30.4\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{0.2}{0.05}=4\left(M\right)\)

\(m_{Fe}=\dfrac{11}{56}=0.19\left(mol\right)\)

\(Fe+H_2SO_4->FeSO_4+H_2\)

0.19   0.19            0.19              (mol)

\(m_{FeSO_4}=0.19\cdot152=28.28\left(g\right)\)

\(Cm=\dfrac{0.19}{0.05}=3.8M\)

nFe=11.256=0.2(mol)

mFeSO4=0,2⋅152=30,4(g)

\(a)n_{CuO}=\dfrac{8}{80}=0,1mol\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{CuCl_2}=n_{Cu}=0,1mol\\ m_{CuCl_2}=0,1.135=13,5g\\ b)n_{HCl}=0,1.2=0,2mol\\ C_{M_{HCl}}=\dfrac{0,2}{0,2}=1M\\ c)C_{M_{CuCl_2}}=\dfrac{0,1}{0,2}=0,5M\)

Phản ứng thuộc phản ứng oxh- khử
VHCl= 400ml => VHCl= 0,4 lít
    mZn= 1.3g =>  nZn= mZn/MZn =  1.3/65 = 0.02(mol)
         Zn + 2HCl ---> ZnCl2 + H2
         1         2             1        1
mol: 0.02    0.04         0.02    0.02
a) mZnCl2 = nZnCl2 . MZnCl2 = (0.02)( 65+ 35,5.2) =2,72g
b) VH2= nH2. 22,4 = 0.02 . 22.4 = 0.448(lít)
c) CM(HCl)= nHCl/VHCl = 0.04/0.4 = 0,1 M