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

$\text{Câu 1:}$

$\text{a)}$

$Fe + 2HCl→FeCl_{2} + H_{2}$

$\text{b)}$

$\text{$n_{H_{2}}$=$\dfrac{V}{22,4}$=$\dfrac{11,2}{22,4}$= 0,5 (mol)}$

$\text{Theo pt: $n_{Fe}$=$n_{H_{2}}$= 0,5 (mol)}$

$\text{→ $m_{Fe}$=n × M=0,5 × 56= 28 (g)}$

$\text{c)}$

$\text{Theo pt: $n_{HCl}$=2$n_{H_{2}}$= 1 (mol)}$

$\text{→ $V_{HCl}$=$\dfrac{n}{C_{M}}$=$\dfrac{1}{0,5}$= 2 }$ `(l)` = 2000 $ml$

$\text{Câu 2:}$

$\text{a)}$

$Fe + 2HCl→FeCl_{2} + H_{2}$

$\text{b)}$

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

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

$\text{→ $m_{Mg}$=n × M=0,3 × 24= 7,2 (g)}$

$\text{c)}$

$\text{Theo pt: $n_{HCl}$=2$n_{H_{2}}$= 0,6 (mol)}$

$\text{→ $V_{HCl}$=$\dfrac{n}{C_{M}}$=$\dfrac{0,6}{0,5}$= 1,2 }$ `(l)` = 1200 $ml$

Đáp án:

 `↓`

Giải thích các bước giải:

 `(1)`

`n_{H_2}` = `{11,2}/{22,4}` = `0,5` `mol`

`a,` `Fe` + `2HCl` → `FeCl_2` + `H_2` 

`b,` `n_{Fe}` = `n_{H_2}` = `0,5` `mol`

`m_{Fe}` = `0,5` . `56` = `28` `g`

`c,` `n_{HCl}` = `2``n_{H_2}` = `1` `mol`

`V_{ddHCl}` = `1/{0,5}` = `2` `l`

 `(2)`

`n_{H_2}` = `{6,72}/{22,4}` = `0,3` `mol`

`a,` `Mg` + `2HCl` → `MgCl_2` + `H_2` 

`b,` `n_{Mg}` = `n_{H_2}` = `0,3` `mol`

`m_{Mg}` = `0,3` . `24` = `7,2` `g`

`c,` `n_{HCl}` = `2``n_{H_2}` = `0,6` `mol`

`V_{ddHCl}` = `0,6/{0,5}` = `1,2` `l`