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

$\text{Câu 1:}$

$\text{CTDC oxit: RO}$

$\text{RO + $H_{2}$$\xrightarrow{t^o}$R + $H_{2}O$}$

$\text{Ta có: $n_{RO}$=$\dfrac{8}{M_{R} + 16}$ (mol)}$

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

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

$\text{⇔ $\dfrac{8}{M_{R} + 16}$ = 0,1}$

$\text{⇔ 0,1($M_{R}$ + 16) = 8}$

$\text{⇔ 0,1$M_{R}$ + 1,6 = 8}$

$\text{⇔ 0,1$M_{R}$=8 - 1,6= 6,4}$

$\text{⇒ $M_{R}$=$\dfrac{6,4}{0,1}$= 64 (g/mol)}$

$\text{Vậy R là kim loại Cu có hóa trị II}$

$\text{⇒ Chọn A}$

$\text{Câu 2:}$

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

$\text{Ta có: $n_{H_{2}}$=$\dfrac{2,8}{22,4}$= 0,125 (mol)}$

$\text{Theo PT ta có: $n_{O_{2}}$=$\dfrac{1}{2}$$n_{H_{2}}$=$\dfrac{1}{2}$ × 0,125= 0,0625 (mol)}$

$\text{⇒ $V_{O_{2}}$=0,0625 × 22,4= 1,4 (lít)}$

$\text{⇒ Chọn A}$

$\text{Câu 3:}$

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

$\text{Ta có: $n_{Fe}$=$\dfrac{5,6}{56}$= 0,1 (mol)}$

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

$\text{⇒ $V_{H_{2}}$=0,1 × 22,4= 2,24 (lít)}$

$\text{⇒ Chọn A}$

1.A 

`RO+ H_2→ R+ H_2O`

`n_(RO} = n_{H_2}`

`<=>8/{R+16} = {2,24}/{22,4}`

`→ R= 64`$(g/mol)$

`→ R: Cu` 

2. A 

`2H_2+ O_2→ 2H_2O`

`V_{O_2} = 1/2V_{H_2} = 1,4(l)`

3. A 

`FeO+ H_2→ Fe+ H_2O`

`n_{H_2} = n_{Fe} = {5,6}/{56} = 0,1(mol)`

`V= 0,1 .22,4= 2,24(l)`