\(4.\\ a/n_{Al}=\dfrac{5,4}{27}=0,2mol\\2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(0,2......0,3............0,1...............0,3\)

\(V_{H_2}=0,3.24,79=7,437l\\ b.C_{\%H_2SO_4}=\dfrac{0,3.98}{150}\cdot100=19,6\%\\ c.m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)

\(7.\\a) n_{H_2SO_4}=\dfrac{100.9,8}{100.98}=0,1mol\\ ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ n_{ZnO}=n_{ZnSO_4}=n_{H_2SO_4}=0,1mol\\ m_{ZnO}=0,1.81=8,1g\\ b)C_{\%ZnSO_4}=\dfrac{0,1.161}{8,1+100}\cdot100=14,8\%\)

\(9.\\ a)n_{H_2}=\dfrac{2,479}{24,79}=0,1mol\\ Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0,1.....0,2.........0,1........0,1\)

\(m_{Mg}=0,1.24=2,4g\\ m_{MgO}=8,4-2,4=6g\\ b)n_{MgO}=\dfrac{6}{40}=0,15mol\\ MgO+2HCl\rightarrow MgCl_2+H_2O\\ n_{HCl}=0,15.2+0,2=0,5mol\\ m_{ddHCl}=\dfrac{0,5.36,5}{3,65}\cdot100=500g\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ 0,2........0,3...........0,1...........0,3\left(mol\right)\\ a.C_{MddH_2SO_4}=\dfrac{0,3}{0,3}=1\left(M\right)\\ b.m_{Al_2\left(SO_4\right)_3}=342.0,1=34,2\left(g\right)\\ c.V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)

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

a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,3}{0,3}=1M\)

b) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=n.M=0,1.342=34,2\left(g\right)\)

c) \(n_{H_2}=n_{H_2SO_4}=0,3\left(mol\right)\)

\(\Rightarrow V_{H_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)

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

PTHH :

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

0,1        0,1             0,1             0,1 

\(a,m_{MgSO_4}=0,1.120=12\left(g\right)\)

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

\(c,m_{ddH_2SO_4}=\dfrac{0,1.98.100}{10}=98\left(g\right)\)

a.\(n_{Mg}=\dfrac{4,8}{24}=0,2mol\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,2                                     0,2      ( mol )

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

b.\(n_{Na}=\dfrac{6,9}{23}=0,3mol\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

0,3                                         0,15   ( mol )

\(V_{H_2}=0,15.24,79=3,7185l\)

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

  0,05                                              0,075      ( mol )

\(V_{H_2}=0,075.24,79=1,85925l\)

Bài 8:

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

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

_____0,2______0,6_____0,2____0,3 (mol)

a, \(m_{Al}=0,2.27=5,4\left(g\right)\)

b, \(C_{M_{HCl}}=\dfrac{0,6}{0,3}=2\left(M\right)\)

c, \(C_{M_{AlCl_3}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Bài 9:

Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

a, \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)

\(\Rightarrow m_{MgO}=8,4-2,4=6\left(g\right)\)

b, \(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}+2n_{MgO}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{3,65\%}==500\left(g\right)\)

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

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

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

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

`Fe + H_2 SO_4 -> FeSO_4 + H_2 ↑`

`0,3`        `0,3`               `0,3`       `0,3`       `(mol)`

`n_[Fe] = [ 16,8 ] / 56 = 0,3 (mol)`

`a) m_[dd H_2 SO_4] = [ 0,3 . 98 ] / [ 9,8 ] . 100 = 300 (g)`

`b) V_[H_2] = 0,3 . 22,4 = 6,72 (l)`

`c) C%_[FeSO_4] = [ 0,3 . 152 ] / [ 16,8 + 300 - 0,3 . 2 ] . 100 ~~ 14,42%`