Bài 1:

a, PTHH: $\text{$Fe_{2}$$O_{3}$ + 3CO $\longrightarrow$ 2Fe + $3CO_{2}$ }$

b,

Ta có: $\text{$n_{CO}$ = $\frac{1,456}{22,4}$ = 0,065 mol}$

Theo PTHH $\rightarrow$ $\text{$n_{Fe_{2}O_{3}}$ = $\dfrac{1}{3}$. $n_{CO}$ =0,02 mol}$

$\rightarrow$ $\text{$m_{Fe_{2}O_{3}}$ = n. M = 0,02. 160 = 3,2 gam}$

c,

Theo PTHH $\rightarrow$ $\text{$n_{CO_{2}}$ = $n_{CO}$ =0,065 mol}$

$\rightarrow$ $\text{$V_{CO_{2}(đktc)}$ = n. 22,4 = 0,065. 22,4 = 1,456 lít}$

Bài 2:

a, PTHH: $\text{Fe + 2HCl $\longrightarrow$ $FeCl_{2}$ + $H_{2}$}$ 

b, Ta có: $\text{$n_{Fe}$ = $\dfrac{m}{M}$ = $\dfrac{5,6}{56}$ = 0,1 mol}$

            $\text{$n_{HCl}$ = $\dfrac{m}{M}$ = $\dfrac{3,65}{36,5}$ = 0,1 mol}$

Ta có tỷ lệ: $\text{$\dfrac{n_{Fe}}{1}$ = $\dfrac{0,1}{1}$ > $\dfrac{n_{HCl}}{2}$ = $\dfrac{0,1}{2}$}$

$\rightarrow$ $\text{Fe dư, HCl hết}$

$\rightarrow$ $\text{$n_{Fe}(p.ư)$ = $\dfrac{0,1}{2}$ = 0,05 mol}$

$\rightarrow$ $\text{$n_{Fe(dư)}$ = 0,1- 0,05 = 0,05 mol }$

$\rightarrow$ $\text{$m_{Fe(dư)}$ = n. M = 0,05. 56 = 2,8 gam }$

c, Theo PTHH $\rightarrow$ $\text{$n_{H_{2}}$ = $\dfrac{1}{2}$. $n_{HCl}$ = 0,05 mol}$

$\rightarrow$ $\text{$V_{H_{2}(đktc)}$ = n. 22,4}$

$\rightarrow$ $\text{$V_{H_{2}(đktc)}$ = 0,05. 22,4= 1,12 lít}$

$\text{#Kudo}$