Ta có: \(n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)

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

               0,4 <---  0,6    ----------->      0,2    -->   0,6

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4.27=10,8\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\\V_{H_2}=0,6.22,4=13,44\left(lít\right)\end{matrix}\right.\)

a) $n_{H_2SO_4} = \dfrac{44,1}{98} = 0,45(mol)$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$

Theo PTHH : 

$n_{Al} = \dfrac{2}{3}n_{H_2SO_4} = 0,3(mol)$
$m_{Al} = 0,3.27 = 8,1(gam)$

b) $n_{H_2} = n_{H_2SO_4} = 0,45(mol)$
$\Rightarrow V_{H_2} = 0,45.22,4 =1 0,08(lít)$

c)

Cách 1 : $n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = 0,15(mol)$
$\Rightarrow m_{Al_2(SO_4)_3} = 0,15.342 = 51,3(gam)$

Cách 2 : Bảo toàn khối lượng, $m_{Al_2(SO_4)_3} = 8,1 + 44,1 - 0,45.2 = 51,3(gam)$

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\)

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

0,6       1,2           0,6        0,6    ( mol )

\(m_{Fe}=0,6.56=33,6g\)

\(m_{FeCl_2}=0,6.127=76,2g\)

\(C_{M_{HCl}}=\dfrac{1,2}{0,6}=2M\)

cái nào a cái nào b ta?

`Fe + 2HCl -> FeCl_2 + H_2↑`

`0,3`    `0,6`         `0,3`       `0,3`     `(mol)`

`n_[H_2] = [ 6,72 ] / [ 22,4 ] = 0,3 (mol)`

  `-> m_[Fe] = 0,3 . 56 = 16,8 (g)`

  `-> m_[FeCl_2] = 0,3 . 127 = 38,1 (g)`

`b) C_[M_[HCl]] = [ 0,6 ] / [ 0,3 ] = 2 (M)`

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂ 

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

=> \(\left\{{}\begin{matrix}m_{Fe}=0,3.56=16,8\left(g\right)\\m_{FeCl_2}=127.0,3=38,1\left(g\right)\\C_{M\left(HCl\right)}=\dfrac{0,6}{0,3}=2M\end{matrix}\right.\)

Theo gt ta có: $n_{Al}=0,2(mol)$

$2Al+6HCl\rightarrow 2AlCl_3+3H_2$

a, Ta có: $n_{H_2}=0,3(mol)\Rightarrow V_{H_2}=6,72(l)$

b, Ta có: $n_{HCl}=0,6(mol)\Rightarrow \%C_{HCl}=10,95\%$

c, Sau phản ứng dung dịch chứa 0,2 mol AlCl3

Suy ra $\%C_{AlCl_3}=13,03\%$

\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\\ n_{H_2SO_4}=\dfrac{49}{98}=0.5\left(mol\right)\)

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

\(Bđ:0.2..........0.5\)

\(Pư:0.2.........0.3..............0.1............0.3\)

\(Kt:0...........0.2...............0.1.............0.3\)

\(m_{Al_2\left(SO_4\right)_3}=0.1\cdot342=34.2\left(g\right)\)

\(V_{H_2}0.3\cdot22.4=6.72\left(l\right)\)

nAl = 5.4/27 = 0.2 mol 

2Al + 3H2SO4 => Al2(SO4)3 + 3H2 

0.2______________0.1________0.3 

VH2 = 0.3*22.4 = 6.72 (l) 

mAl2(SO4)3 = 0.1*342 = 34.2 g 

a) $2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$

b)

$n_{H_2SO_4} = \dfrac{200.15\%}{98} = \dfrac{15}{49}(mol)$

Theo PTHH :

$n_{H_2} = n_{H_2SO_4} = \dfrac{15}{49}(mol)$
$n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = \dfrac{5}{49}(mol)$

Vậy :

$V_{H_2} = \dfrac{15}{49}.22,4 = 6,86(lít)$
$m_{Al_2(SO_4)_3} = \dfrac{5}{49}.342 = 34,9(gam)$

\(n_{H_2SO_4}=\dfrac{200\cdot15\%}{98}=\dfrac{15}{49}\left(mol\right)\)

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

\(........\dfrac{15}{49}.........\dfrac{5}{49}......\dfrac{15}{49}\)

\(m_{Al_2\left(SO_4\right)_3}=\dfrac{5}{49}\cdot342=35\left(g\right)\)

\(V_{H_2}=\dfrac{15}{49}\cdot22.4=6.85\left(l\right)\)

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

Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,15\left(mol\right)=n_{H_2}\\n_{Al_2\left(SO_4\right)_3}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.\)

a) \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

______0,1--->0,15-------->0,05------->0,15

=> m_H2SO4 = 0,15.98 = 14,7 (g)

b) V_H2 = 0,15.22,4 = 3,36 (l)

c) m_Al2(SO4)3 = 0,05.342 = 17,1 (g)