Trong \(20,4g\) hỗn hợp có: \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\\n_{Al}=c\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow65a+56b+27c=20,4\left(1\right)\)

\(n_{H_2}=\dfrac{10,08}{22,4}=0,45mol\)

\(BTe:2n_{Zn}+2n_{Fe}+3n_{Al}=2n_{H_2}\)

\(\Rightarrow2a+2b+3c=2\cdot0,45\left(2\right)\)

Trong \(0,2mol\) hhX có \(\left\{{}\begin{matrix}Zn:ka\left(mol\right)\\Fe:kb\left(mol\right)\\Al:kc\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow ka+kb+kc=0,2\)

\(n_{Cl_2}=\dfrac{6,16}{22,4}=0,275mol\)

\(BTe:2n_{Zn}+3n_{Fe}+3n_{Al}=2n_{Cl_2}\)

\(\Rightarrow2ka+3kb+3kc=2\cdot0,275\)

Xét thương:

 \(\dfrac{ka+kb+kc}{2ka+3kb+3kc}=\dfrac{0,2}{2\cdot0,275}\Rightarrow\dfrac{a+b+c}{2a+3b+3c}=\dfrac{4}{11}\)

\(\Rightarrow3a-b-c=0\left(3\right)\)

Từ (1), (2), (3)\(\Rightarrow\left\{{}\begin{matrix}a=0,1mol\\b=0,2mol\\c=0,1mol\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Zn}=6,5g\\m_{Fe}=11,2g\\m_{Al}=2,7g\end{matrix}\right.\)

Vì R là kim loại kiềm nên R thuộc nhóm IA

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

\(2R+H_2O->R_2O+H_2\)

0,6←                            0,3 (mol)

⇒ \(M_R=\frac{m_R}{n_R}=\frac{13,8}{0,6}=23 (g/mol)\)

⇒ R là natri (Na)

2M + 6 HCl -> 2 MCl2 + 3 H2

nH2=10,08/22,4=0,45(mol)

=>nM=0,3(mol)

=>M(M)=8,1/0,3=27(g/mol)

=>M(III) cần tìm là nhôm (Al=27)

n_H2 = 10.08/22.4 = 0.45 (mol) 

2M + 6HCl => 2MCl₃ + 3H₂

0.9...............................0.45 

M_M = 8.1/0.3 =  27

M là : Al

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

PTHH: R + 2HCl → RCl₂ + H₂

Mol:   0,15     0,3                 0,15

\(M_R=\dfrac{8,4}{0,15}=56\left(g/mol\right)\)

 ⇒ R là sắt (Fe)

b, \(m_{ddHCl}=\dfrac{0,3.36,5.100}{15}=73\left(g\right)\)

Câu 1: 

A₂O₃ + 3H₂ -t⁰-> 2A + 3H₂O

2A+48...................2A

16..........................11.2 

<=> 11.2 * (2A + 48) = 16 * 2A 

=> A = 56 

Vậy A là : Fe

n_H2 = 10.08/22.4 = 0.45 (mol) 

2M + 6HCl => 2MCl₃ + 3H₂

0.9...............................0.45 

M_M = 8.1/0.3 =  27

M là : Al

1.

3H2+A2O3----t°--}2A+3H2O

Gọi nH2=nH2O=a mol 

Áp dụng định luật bảo toàn khối lượng,ta có 

2a+16=11,2+18a

16a=4,8

a=0,3(mol)

Theo pt:

nA=2/3.nH2=2/3.0.3=0,2(mol)

MA=11,2/0,2=56(g/mol)

A Là Zn

\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(R+2HCl\rightarrow RCl_2+H_2\)

\(0.1........0.2................0.1\)

\(M_R=\dfrac{13.7}{0.1}=137\left(\dfrac{g}{mol}\right)\)

\(R:Ba\)

\(200\left(ml\right)=0.2\left(l\right)\)

\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)