nH2 = 3,36/22,4 = 0,15 (mol)

PTHH: 2R + 6HCl -> 2RCl3 + 3H2

nR = nRCl3 = 0,15 : 2/3 = 0,1 (mol)

M(R) = 2,7/0,1 = 27 (g/mol)

=> R là Al

CMAlCl3 = 0,1/0,2 = 0,5M

Ta có :

2Al + 6HCl ---> 2ACl3 + 3H2

nH2 = 0,15 mol

=> nAl = 0,1 mol

=> M = 2,7/0,1 = 27

=> ĐÓ là Al

nACl3 = 2/3nH2 = 0,1 mol

Cm = n/V = 0,1 / 0,2 = 0,5M

\(n_{HCl}=\dfrac{200\cdot7.3\%}{36.5}=0.4\left(mol\right)\)

\(M+2HCl\rightarrow MCl_2+H_2\)

\(0.2.....0.4.........0.2........0.2\)

\(m_{MCl_2}=0.2\cdot\left(M+71\right)\left(g\right)\)

\(m_{dd}=0.2M+200-0.2\cdot2=0.2M+199.6\left(g\right)\)

\(C\%MCl_2=\dfrac{0.2\cdot\left(M+71\right)}{0.2M+199.6}\cdot100\%=12.05\%\)

\(\Rightarrow M=56\)

\(M:Sắt\)

câu a

dạ em làm xong câu B rồi mọi người khỏi cần trả lời nữa ạ

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

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

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

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

\(R:Ba\)

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

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

a) Gọi kim loại cần tìm là R

\(n_R=\dfrac{7,56}{M_R}\left(mol\right)\)

PTHH: 2R + 2nHCl --> 2RCl_n + nH₂

         \(\dfrac{7,56}{M_R}\)------------>\(\dfrac{7,56}{M_R}\)

=> \(M_{RCl_n}=M_R+35,5n=\dfrac{37,38}{\dfrac{7,56}{M_R}}\)

=> \(M_R=9n\left(g/mol\right)\)

Xét n = 1 => M_R = 9(Loại)

Xét n = 2 => M_R = 18 (Loại)

Xét n = 3 => M_R = 27(g/mol) => R là Al (Nhôm)

b) 

\(n_{Al}=\dfrac{7,56}{27}=0,28\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

          0,28-->0,84--->0,28--->0,42

=> \(V_{H_2}=0,42.22,4=9,408\left(l\right)\)

\(m_{HCl}=0,84.36,5=30,66\left(g\right)\)

=> \(m_{ddHCl}=\dfrac{30,66.100}{12}=255,5\left(g\right)\)

c) m_{dd sau pư} = 7,56 + 255,5 - 0,42.2 = 262,22 (g)

=> \(C\%_{AlCl_3}=\dfrac{37,38}{262,22}.100\%=14,255\%\)

2A+6HCl->2ACl₃+3H₂

0,2----0,6------------0,3 mol

n _H2=\(\dfrac{6,72}{22,4}\)=0,3 mol

=>\(\dfrac{5,4}{A}\)=0,2

=>A=27 g\mol

=>A là nhôm (Al)

CM_HCl=\(\dfrac{0,6}{0,5}\)=1,2M

\(5,4gA+500mlHCl->X:ACl3+6,72lH2\)

nH2 = 0,3 ( mol )

=> nAl = 2/3.nH2 = 0,2 ( mol )

( Cân bằng PTHH )

Ta có :

M = \(\dfrac{5,4}{0,2}=27\left(\dfrac{g}{mol}\right)\)

=>  Đó là Al

1/ PT : X +  2H₂O -> X[OH]₂ + H₂

_{mol :    \(\frac{6}{M_X}\)}             ->                   \(\frac{6}{M_X}\)     

=> m_H2 = \(\frac{12}{M_X}\)       => m_dd = m+6 - \(\frac{12}{M_X}\)

Ta có: m+5,7 = m+6 - \(\frac{12}{M_X}\)

<=> \(\frac{12}{M_X}\)= 0,3 => M_{X = 40 => X là Canxi [Ca]}

2/ Dặt n_HCl= a [a> 0] => m_ddHCl= 36,5a : 14,6 x 100= 250a

     PT : X  + 2HCL => XCl₂ +  H₂

   mol :   a/2        a         ->  a/2        a/2

m_H2 = a/2 x 2 = a ; m_X = a/2 . M_X

m XCl₂= a/2 x [M_X +71]

m_dd XCL2= a/2 .M_X + 250a  - a = a/2 .M_X +249a

Ta có :\(\frac{\frac{a}{2}\times M_X+\frac{71}{2}a}{M_X\times a:2+249a}\times100\%=24,15\%\)

<=> \(\frac{M_X+71}{M_X+498}=24,15\%\Leftrightarrow M_X=65\)=> X là kẽm [Zn]

BTKL: m_D + m_NaHCO3 = m_CO2 + m_E

m_D + 179,88 = 44.0,2 + 492 => m_D = 320,92

BTKL: m_Mg + m_ddHCl = m_H2  + m_D

=> 24 . 0,4 + m_ddHCl = 2 . 0,4 + 320,92 => m_ddHCl = 312,12

=> C%HCl = 11,69%

\(n_Y=\dfrac{6,72}{22,4}=0,3mol\)

\(2R+2HCl\rightarrow2RCl+H_2\)

0,6      0,6         0,6        0,3

Mà \(n_R=\dfrac{23,4}{M_R}=0,6\Rightarrow M_R=39\left(K\right)\)

\(m_{HCl}=0,6\cdot36,5=21,9\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{21,9}{25}\cdot100=87,6\left(g\right)\)

\(m_{H_2}=0,3\cdot2=0,6\left(g\right)\)

\(m_{KCl\left(lt\right)}=0,6\cdot74,5=44,7\left(g\right)\)

\(m_{ddsau}=23,4+87,6-0,6=110,4\left(g\right)\)

\(\Rightarrow C\%=\dfrac{44,7}{110,4}\cdot100=40,5\%\)