`a)`

`Fe + 2HCl -> FeCl_2 +H_2`

`b)`

`n_{Fe} = 12/56 = 3/14 (mol)`

`n_{H_2} = n_{Fe} = 3/14 (mol)`

`=>` `V_{H_2} = 3/4.22,4 = 4,8l`

`c)`

`n_{HCl} = 2n_{Fe} = 3/7 (mol)`

`C_{M\ HCl} = 3/7 : 0,25 = 1,7M`

`d)`

`n_{CuO} = 48/80 = 0,6 (mol)`

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

`n_{CuO} > n_{H_2} -> CuO` dư

`n_{CuO(dư)} = 0,6 - 3/14 = 27/70 (mol)`

`m_{CuO(dư)} = 27/70.80 = 30,86g`

Giải thích các bước giải:

$a)$

$Fe+2HCl\to FeCl_2+H_2$

$b)$

$n_{Fe}=\dfrac{12}{56}=\dfrac{3}{14}(mol)$

$n_{H_2}=n_{Fe}=\dfrac{3}{14}(mol)$

$V_{H_2}=\dfrac{3}{14}×22,4=4,8(l)$

$c)$

$n_{HCl}=2n_{Fe}=\dfrac{3}{7}(mol)$

$C_{M\ HCl}=\dfrac{\dfrac{3}{7}}{0,25}=1,7(M)$

$d)$

$n_{CuO}=\dfrac{48}{80}=0,6(mol)$

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

$n_{CuO}>n_{H_2}\to CuO$ dư.

$n_{CuO\ \rm (dư)}=0,6-\dfrac{3}{14}=\dfrac{27}{70}(mol)$

$m_{CuO\ \rm (dư)}=\dfrac{27}{70}×80=30,86(g)$