Đáp án+Giải thích các bước giải:

$\text{Câu 8:}$

$2X + 2H_{2}O→2XOH + H_{2}$

$\text{$n_{X}$=$\dfrac{11,7}{M_{X}}$ (mol)}$

$\text{$n_{H_{2}}$=$\dfrac{3,36}{22,4}$= 0,15 (mol)}$

$\text{Theo pt: $n_{X}$=2$n_{H_{2}}$=2 × 0,15= 0,3 (mol)}$

$\text{⇔ $\dfrac{11,7}{M_{X}}$ = 0,3}$

$\text{⇒ $M_{X}$=$\dfrac{11,7}{0,3}$= 39 (g/mol)}$

Vậy $X$ là kim loại $K$ có hóa trị $I$

$\text{Câu 9:}$

$2M + 2H_{2}O→2MOH + H_{2}$

$\text{$n_{M}$=$\dfrac{9,2}{M_{M}}$ (mol)}$

$\text{$n_{H_{2}}$=$\dfrac{4,48}{22,4}$= 0,2 (mol)}$

$\text{Theo pt: $n_{M}$=2$n_{H_{2}}$=2 × 0,2= 0,4 (mol)}$

$\text{⇔ $\dfrac{9,2}{M_{M}}$ = 0,4}$

$\text{⇔ $M_{M}$=$\dfrac{9,2}{0,4}$= 23 (g/mol)}$

Vậy $M$ là kim loại $Na$ có hóa trị $I$

$\text{Câu 10:}$

$X + 2H_{2}O→X(OH)_{2} + H_{2}$

$\text{$n_{X}$=$\dfrac{12}{M_{X}}$ (mol)}$

$\text{$n_{H_{2}}$=$\dfrac{6,72}{22,4}$= 0,3 (mol)}$

$\text{Theo pt: $n_{X}$=$n_{H_{2}}$= 0,3 (mol)}$

$\text{⇔ $\dfrac{12}{M_{X}}$ = 0,3}$

$\text{⇒ $M_{X}$=$\dfrac{12}{0,3}$= 40 (g/mol)}$

Vậy $X$ là kim loại $Ca$ có hóa trị $II$

Em tham khảo nha:

\(\begin{array}{l}
8)\\
2X + 2{H_2}O \to 2XOH + {H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_X} = 2{n_{{H_2}}} = 0,3\,mol\\
{M_X} = \dfrac{{11,7}}{{0,3}} = 39g/mol \Rightarrow X:Kali(K)\\
9)\\
2M + 2{H_2}O \to 2MOH + {H_2}\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_M} = 2{n_{{H_2}}} = 0,4\,mol\\
{M_M} = \dfrac{{9,2}}{{0,4}} = 23g/mol \Rightarrow M:Natri(Na)\\
10)\\
X + 2{H_2}O \to X{(OH)_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_X} = {n_{{H_2}}} = 0,3\,mol\\
{M_X} = \dfrac{{12}}{{0,3}} = 40g/mol \Rightarrow X:Canxi(Ca)
\end{array}\)