\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ PTHH:Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow n_{H_2SO_4}=n_{H_2}=0,15(mol)\\ \Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,15}{0,05}=3M\\ PTHH:2H_2+O_2\xrightarrow{t^o}2H_2O\\ \Rightarrow n_{O_2}=\dfrac{1}{2}n_{H_2}=0,075(mol)\\ \Rightarrow V_{O_2}=0,075.22,4=1,68(l)\)

Fe+2HCl->FeCl2+H2

0,3----0,6--0,3------0,3

n H2=0,3 mol

=>m Fe=0,3.56=16.8g

=>CM=0,6\0,1=6M

CuO+H2-tO>Cu+H2O

0.3---------------0,3

=>m Cu=0,3.64=19,2g

câu 2

FE+H2SO4->FeSO4+H2

0,25----0,25----0,25---0,25

n Fe=0,25 mol

=>mFeCl2=0,25.127=31,75g

=>VH2=0,25.22,4=5,6l

=>C% muối=\(\dfrac{31,75}{14+400-0,25.2}100=7,67\%\)

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

nH2= 0,15(mol)

=> nZn=nH2SO4=nZnSO4=nH2=0,15(mol)

b) mZn=0,15.65=9,75(g)

c) CMddH2SO4= 0,15/ 0,05=3(M)

d) mZnSO4= 161. 0,15=24,15(g)

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

\(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)

b, \(n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{0,3.98}{120}.100\%=24,5\%\)

c, m _{dd sau pư} = 16,8 + 120 - 0,3.2 = 136,2 (g)

d, \(n_{FeSO_4}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow C\%_{FeSO_4}=\dfrac{0,3.152}{136,2}.100\%\approx33,48\%\)

a, \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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

\(n_{Fe}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(n_{NaOH}=2n_{Na_2O}=0,2\left(mol\right)\Rightarrow C_{M_{NaOH}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)

\(a,n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 

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

0,15    0,3           0,15      0,15

\(m_{Fe}=0,15.56=8,4\left(g\right)\)

\(a,n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

PTHH :

\(Na_2O+H_2O\rightarrow2NaOH\)

0,1           0,1        0,2

\(C_{M\left(A\right)}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)

a) Fe+H2SO4------>FeSO4+H2

b)nH2=5,6/22,4=0,25mol

Fe+H2SO4------->FeSO4+H2

0,25 <----                           0,25

Theo pt:

nFe=0,25mol

=>mFe=0,25×56=14g

c)

Fe+H2SO4------->FeSO4+H2

         0,25 <----                  0,25

Theo pt:

nH2SO4=0,25 mol

CM=n/V=0,25/0,05=5(M)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

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

0,25 .... 0,25 ...... 0,25 ..... 0,25 (mol) 

a. \(m_{Fe}pư=0,25.56=14\left(g\right)\)

b. \(C_M\left(H_2SO_4\right)=C_M\left(FeSO_4\right)=\dfrac{0,25}{0,05}=5\left(M\right)\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)\\ Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow n_{Fe}=n_{H_2SO_4}=n_{Fe(OH)_2}=0,2(mol)\\ a,m_{Fe}=0,2.56=11,2(g)\\ b,C_{M_{H_2SO_4}}=\dfrac{0,2}{0,2}=1M\)

\(c,Ba(OH)_2+FeSO_4\to BaSO_4\downarrow+Fe(OH)_2\downarrow\\ n_{Ba(OH)_2}=\dfrac{250.17,1}{100.171}=0,25(mol)\\ LTL:\dfrac{0,2}{1}<\dfrac{0,25}{1}\Rightarrow Ba(OH)_2\text{ dư}\\ \Rightarrow n_{BaSO_4}=0,2(mol)\\ \Rightarrow m_{BaSO_4}=233.0,2=46,6(g)\)

a) \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: Mg + H₂SO₄ --> MgSO₄ + H₂

          0,1<----------------0,1<----0,1

=> m_Mg = 0,1.24 = 2,4 (g)

=> m = m_Cu = 10,2 - 2,4 = 7,8 (g)

b) 

\(C_{M\left(MgSO_4\right)}=\dfrac{0,1}{0,5}=0,2M\)