\(n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2} = \dfrac{3}{2}n_{Al} = 0,15(mol)\\ \Rightarrow V_{H_2} = 0,15.22,4 = 3,36(lít)\\ n_{HCl} = \dfrac{8,1}{36,5} = \dfrac{81}{365}(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ \dfrac{n_{Al}}{2} = 0,05 > \dfrac{n_{HCl}}{6} = \dfrac{27}{730} \to Al\ dư\\ n_{Al\ pư} = \dfrac{1}{3}n_{HCl} = \dfrac{27}{365}(mol)\\ \)

\(m_{Al\ dư} = 2,7 - \dfrac{27}{265}.27 = 0,703(gam)\)

a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{54}{27}=2\left(mol\right)\)

Theo PTHH: \(n_{H_2}=\dfrac{2.3}{2}=3\left(mol\right)\)

\(\Rightarrow V_{H_2}=n_{H_2}.22,4=3.22,4=67,2l\)

b) \(2H_2+O_2\rightarrow2H_2O\)

\(n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{30}{32}=0,94\left(mol\right)\)

Theo PTHH: \(n_{H_2O}=\dfrac{0,94.2}{1}=1,88\left(mol\right)\)

\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=1,88.18=33,84\left(g\right)\)

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

       0,15<--0,3<-----0,15<--0,15

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

=> \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)

b) \(m_{FeCl_2}=0,15.127=19,05\left(g\right)\)

a, \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)

b, \(n_{CH_3COOH}=\dfrac{3,6}{60}=0,06\left(mol\right)\)

Theo PT: \(n_{\left(CH_3COO\right)_2Mg}=n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=0,03\left(mol\right)\)

\(\Rightarrow m_{\left(CH_3COO\right)_2Mg}=0,03.142=4,26\left(g\right)\)

\(V_{H_2}=0,06.22,4=1,344\left(l\right)\)

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

\(n_{Al}=\dfrac{3,375}{27}=0,125\)

\(\Rightarrow n_{H_2}=\dfrac{0,125}{2}.3=0,1875mol\)      \(\Rightarrow V_{H_2}=0,1875.22,4=4,2l\)

\(n_{AlCl_3}=n_{Al}=0,125mol\)       \(\Rightarrow m_{AlCl_3}=0,125.133,5=16,6875g\)

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 

\(nAl=\dfrac{13,5}{27}=0,5\left(mol\right)\)

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

 2          6             2            3    (mol)

0,5     1,5          0,5           0,75   (mol)

a. \(VH_2=0,75.22,4=16,8\left(l\right)\)

b.

 \(FeO+H_2\rightarrow Fe+H_2O\)

   1        1        1           1    (mol)

 0,75    0,75   0,75    0,75

\(m_{Fe}=0,75.42\left(g\right)\)

gọi kim loại hóa trị 2 là A.

Số mol của H có trong 1,2 g H₂ là: n=1,2/2=0,6 mol

SĐPƯ: A + 2HCL ------ACL_{2 ⁺} H₂

0,6mol 1,2mol 0,6 mol

a, khối lượng HCL đã phản ứng là: m= 1,2 * 36,5= 43,8 g

b, số mol kim loại A là 0,6 mol

công thức của kim loại A là : 32,5 / 0,6 = \(\frac{32,5}{0,6}\approx55\)

vậy A là mângn

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

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

Theo PTHH: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)

Theo PTHH: \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=127\cdot0,1=12,7\left(g\right)\)