$\text{Đáp án+ Giải thích các bước giải:}$

Bài 1:

a. $n_{H_2}$= $\frac{13,44}{22,4}$= $0,6(mol)$

   $n_{O_2}$= $\frac{5,6}{22,4}$= $0,25(mol)$

Pt: $2H_2$ + $O_2$ $\xrightarrow{t^o}$ $2H_2O$ $(1)$

$\text{Ta có: $\dfrac{0,6}{2}$ > $\dfrac{0,25}{1}$ }$

$\Rightarrow$ $\text{$H_2$ dư, $O_2$ hết.}$

$\text{Từ pt(1):$n_{H_2O}$ = 0,5(mol) }$

$\Rightarrow$ $m_{H_2O}$ = $0,5× 18= 9(g)$

b. $\text{Khí $H_2$ dư( 0,1)}$

$\Rightarrow$ $V_{H_2 dư}$ = $0,1× 22,4= 2,24()$  

Bài 2: 

$n_{H_2}$= $\frac{22,4}{22,4}$= $1(mol)$

$n_{O_2}$= $\frac{11,2}{22,4}$= $0,5(mol)$

Pt: $2H_2$ + $O_2$ $\xrightarrow{t^o}$ $2H_2O$ $(1)$

$\text{Ta có: $\dfrac{1}{2}$ = $\dfrac{0,5}{1}$ }$

$\Rightarrow$ $\text{Pứ vừa đủ.}$

$\text{Từ pt(1):$n_{H_2O}$ = 1(mol) }$

$\Rightarrow$ $V_{H_2O}$ = $1× 22,4= 22,4(l)$

Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
{n_{{H_2}}} = \dfrac{{13,44}}{{22,4}} = 0,6mol\\
{n_{{O_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
2{H_2} + {O_2} \to 2{H_2}O\\
\dfrac{{0,6}}{2} > \dfrac{{0,25}}{1} \Rightarrow {H_2}\text{ dư}\\
{n_{{H_2}O}} = 2{n_{{O_2}}} = 0,5mol\\
{m_{{H_2}O}} = 0,5 \times 18 = 9g\\
b)\\
{n_{{H_2}d}} = 0,6 - 0,25 \times 2 = 0,1mol\\
{V_{{H_2}d}} = 0,1 \times 22,4 = 2,24l\\
2)\\
{n_{{H_2}}} = \dfrac{{22,4}}{{22,4}} = 1mol\\
{n_{{O_2}}} = \dfrac{{11,2}}{{22,4}} = 0,5mol\\
2{H_2} + {O_2} \to 2{H_2}O\\
\dfrac{1}{2} = \dfrac{{0,5}}{1} \Rightarrow \text{Phản ứng xảy ra hoàn toàn}\\
{n_{{H_2}O}} = {n_{{H_2}}} = 1mol\\
{V_{{H_2}O}} = 1 \times 22,4 = 22,4l
\end{array}\)