Đáp án + Giải thích các bước giải:

Bài 1:

$n_{Fe}=\frac{5,6}{56}=0,1(mol)$

$n_{Fe_2O_3}=\frac{16}{160}=0,1(mol)$

$=>\overline{M}_{\rm hỗn\,hợp}=\frac{5,6+16}{0,1+0,1}=108(g/mol)$

Bài 2:

$a\bigg)$

$n_{Fe}=\frac{5,6}{56}=0,1(mol)$

$Fe+2HCl\to FeCl_2+H_2$

Xét tỉ lệ: $n_{Fe}>\frac{n_{HCl}}{2}=>Fe$ dư

Theo PT: $n_{H_2}=\frac{1}{2}n_{HCl}=0,025(mol)$

$=>V_{H_2(đktc)}=0,025.22,4=0,56(l)$

$b\bigg)$

Theo PT: $n_{Fe(pứ)}=n_{H_2}=0,025(mol)$

$=>m_{Fe(dư)}=(0,1-0,025).56=4,2(g)$

Bài 3:

$a\bigg)$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

$CuO+H_2\xrightarrow{t^o}Cu+H_2O$

$b\bigg)$

Ta có $m_{Fe_2O_3}:m_{CuO}=2:1$

$=>m_{Fe_2O_3}=\frac{2}{3}m_A=\frac{2}{3}.240=160(g)$

$=>m_{CuO}=240-160=80(g)$

$=>\begin{cases} n_{Fe_2O_3}=\frac{160}{160}=1(mol)\\ n_{CuO}=\frac{80}{80}=1(mol) \end{cases}$

Theo PT: $n_{Fe}=2n_{Fe_2O_3}=2(mol);n_{Cu}=n_{CuO}=1(mol)$

$=>\begin{cases} m_{Fe}=2.56=112(g)\\ m_{Cu}=1.64=64(g) \end{cases}$

Đáp án + Giải thích các bước giải:

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

$n_{Fe_2O_3} = \dfrac{m}{M} = \dfrac{16}{2 . 56 + 3 . 16} = \dfrac{16}{160} = 0,1\text{ mol}$
$n_{hh} = 0,1 + 0,1 = 0,2\text{ mol}$

$m_{hh} = 5,6 + 16 = 21,6g$

$\overline{M}_{hh} = \dfrac{m}{n} = \dfrac{21,6}{0,2} = 108g/mol$

$----------------------$

Bài 2:
a) $n_{Fe} = \dfrac{m}{M} = \dfrac{5,6}{56} = 0,1\text{ mol}$
PTHH:
$Fe + 2HCl \longrightarrow FeCl_2 + H_2 \uparrow$Ta có:
$\dfrac{n_{Fe}}{1} > \dfrac{n_{HCl}}{2}\bigg(\dfrac{0,1}{1} > \dfrac{0,05}{2}\bigg)$

$\Rightarrow Fe$ dư, tính theo $HCl$

Theo PTHH:

$n_{H_2} = \dfrac{1}{2}n_{HCl} = \dfrac{1}{2} . 0,05 = 0,025\text{ mol}$

$V_{H_2} = n . 22,4 = 0,025 . 22,4 = 0,56l$

$----------------------$

b) $Fe$ dư

Theo PTHH:

$n_{Fe_{pứ}} = \dfrac{1}{2}n_{HCl} = \dfrac{1}{2} . 0,05 = 0,025\text{ mol}$

$n_{Fe_{\text{dư}}} = 0,1 - 0,025 = 0,075\text{ mol}$

$m_{Fe_{\text{dư}}} = n . M  = 0,075 . 56 = 4,2g$

$----------------------$

Bài 3:
a) PTHH:
$Fe_2O_3 + 3H_2 \overset{t^o}{\longrightarrow} 2Fe + 3H_2O(1)$

$CuO + H_2 \overset{t^o}{\longrightarrow} Cu + H_2O(2)$

$----------------------$

b) $m_{Fe_2O_3} = \dfrac{2}{2 + 1}m_A = \dfrac{2}{3} . 240 = 160g$

$n_{Fe_2O_3} = \dfrac{m}{M} = \dfrac{160}{2 . 56 + 3 . 16} = \dfrac{160}{160} = 1\text{ mol}$

$m_{CuO} = 240 - 160 = 80g$

$n_{CuO} = \dfrac{m}{M} = \dfrac{80}{64 + 16} = \dfrac{80}{80} = 1\text{ mol}$

Theo PTHH(1):

$n_{Fe} = 2n_{Fe_2O_3} = 2\text{ mol}$

$m_{Fe} = n . M = 2 . 56 = 112g$

Theo PTHH(2):

$n_{Cu} = n_{CuO} = 1\text{ mol}$

$m_{Cu} = n . M = 1 . 64 = 64g$