Đáp án B

\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(A+2H_2O\rightarrow A\left(OH\right)_2+H_2\)

\(................0.15.....0.15\)

\(M_{A\left(OH\right)_2}=\dfrac{11.1}{0.15}=74\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow A=40\)

\(\Rightarrow B\)

$A + 2H_2O \to A(OH)_2 + H_2$

n A(OH)2 = n H2 = 3,36/22,4 = 0,15(mol)

M A(OH)2 = A + 34 = 11,1/0,15 = 74

=> A = 40(Ca)

Vậy A là Canxi 

Đáp án B

B.Ca

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\); \(n_{A\left(OH\right)_2}=\dfrac{34,2}{M_A+34}\left(mol\right)\)

\(A+2H_2O\rightarrow A\left(OH\right)_2+H_2\)

                 \(\dfrac{34,2}{M_A+34}\)  --> \(\dfrac{34,2}{M_A+34}\)  ( mol )

\(\rightarrow n_{H_2}=\dfrac{34,2}{M_A+34}=0,2\left(mol\right)\)

\(\Leftrightarrow34,2=0,2M_A+6,8\)

\(\Leftrightarrow0,2M_A=27,4\)

\(\Leftrightarrow M_A=137\) ( g/mol )

--> A là Bari ( Ba )

\(A+H_2O\rightarrow A\left(OH\right)_2+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ TheoPT:n_{H_2}=n_{A\left(OH\right)_2}=0,2\left(mol\right)\\ \Rightarrow M_{A\left(OH\right)_2}=A+17.2=\dfrac{34,2}{0,2}=171\\ \Rightarrow A=137\left(Ba\right)\)

`A + 2H_2 O -> A(OH)_2 + H_2`

`n_[H_2] = [ 4,48 ] / [22,4 ] = 0,2 (mol)`

`n_[A(OH)_2] = [ 34,2 ] / [ M_A + 34 ] (mol)`

Mà `n_[H_2] = n_[A(OH)_2]`

 `=> [34,2] / [M_A + 34] = 0,2`

`<=>M_A = 137 ( g // mol)`

         `-> A` là `Ba`

\(a.R+2H_2O\rightarrow R\left(OH\right)_2+H_2\\ b.n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\\ n_R=n_{H_2}=0,5mol\\ M_R=\dfrac{20}{0,5}=40g/mol\)
Vậy kim loại R là Ca

\(a,PTHH:X+2HCl\to XCl_2+H_2\\ \Rightarrow n_{X}=n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ \Rightarrow M_X=\dfrac{9,75}{0,15}=65(g/mol)(Zn)\\ b,n_{HCl}=2.0,2=0,4(mol)\)

Vì \(\dfrac{n_{H_2}}{1}<\dfrac{n_{HCl}}{2}\) nên \(HCl\) dư

\(\Rightarrow n_{ZnCl_2}=n_{H_2}=0,15(mol)\\ \Rightarrow m_{ZnCl_2}=136.0,15=20,4(g)\\ C_{M_{ZnCl_2}}=\dfrac{0,15}{0,2}=0,75M\)

=> KL M là Kali (M=39, n=1)

Đáp án D