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

$\text{A.}$

$\text{$2H_{2}$ + $O_{2}$$\xrightarrow{t^o}$$2H_{2}O$ (1)}$

$\text{B.}$

$\text{Ta có: $n_{H_{2}}$=$\dfrac{2,24}{22,4}$= 0,1 (mol)}$

$\text{Theo PT (1) ta có: $n_{O_{2}}$=$\dfrac{1}{2}$$n_{H_{2}}$=$\dfrac{1}{2}$ × 0,1= 0,05 (mol)}$

$\text{⇒ $V_{O_{2}}$=0,05 × 22,4= 1,12 (lít)}$

$\text{⇒ $V_{KK}$=1,12 × 5= 5,6 (lít)}$

$\text{C.}$

$\text{FeO + $H_{2}$$\xrightarrow{t^o}$Fe + $H_{2}O$ (2)}$

$\text{Ta có: $n_{FeO}$=$\dfrac{10,8}{72}$= 0,15 (mol)}$

$\text{$n_{H_{2}}$= 0,05 (mol)}$

$\text{Lập tỉ lệ: $\dfrac{0,15}{1}$>$\dfrac{0,05}{1}$}$

$\text{⇒ FeO dư, bài toán theo số mol của $H_{2}$}$

$\text{Theo PT (2) ta có: $n_{Fe}$=$n_{H_{2}}$= 0,05 (mol)}$

$\text{⇒ $m_{Fe}$=0,05 × 56= 2,8 (g)}$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2{H_2} + {O_2} \to 2{H_2}O\\
b)\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{{O_2}}} = \dfrac{{{n_{{H_2}}}}}{2} = 0,05mol\\
{V_{{O_2}}} = 0,05 \times 22,4 = 1,12l\\
{V_{kk}} = 5{V_{{O_2}}} = 5,6l\\
c)\\
FeO + {H_2} \to Fe + {H_2}O\\
{n_{FeO}} = \dfrac{{10,8}}{{72}} = 0,15mol\\
\dfrac{{0,15}}{1} > \dfrac{{0,1}}{1} \Rightarrow FeO\text{ dư}\\
{n_{Fe{O_d}}} = 0,15 - 0,1 = 0,05mol\\
{m_{Fe{O_d}}} = 0,05 \times 72 = 3,6g\\
{n_{Fe}} = {n_{{H_2}}} = 0,1mol\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
{m_{cr}} = 3,6 + 5,6 = 9,2g
\end{array}\)