1) a)

\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
nFe = \frac{{11,2}}{{56}} = 0,2mol\\
nHCl = \frac{{7,3}}{{36,5}} = 0,2\,mol
\end{array}\)

Ta có

\(\frac{{0,2}}{1} > \frac{{0,2}}{2}\)

=> Fe dư

nFe dư=0,1 mol

mFe dư=0,1x56=5,6 g

b)

VH2=0,1x22,4=2,24 l

c)

mFeCl2=0,1x127=12,7 g

2)

a)

\(\begin{array}{l}
CuO + {H_2} \to Cu + {H_2}O\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
b)\\
mCuO = 80 \times 40\%  = 32g\\
nCuO = \frac{{32}}{{80}} = 0,4mol\\
nF{e_2}{O_3} = \frac{{48}}{{160}} = 0,3mol\\
nCu = nCuO = 0,4mol\\
mCu = 0,4 \times 64 = 25,6g\\
nFe = 2nF{e_2}{O_3} = 0,6mol\\
mFe = 0,6 \times 56 = 33,6g
\end{array}\)

Câu 1 :

$n_{Fe}=11,2/56=0,2mol$

$n_{HCl}=7,3/36,5=0,2mol$

$PTHH :$

                        $Fe   +   2HCl\to FeCl_2+H_2↑$

$\text{Theo pt :       1 mol       2 mol}$

$\text{Theo đbài :  0,2 mol    0,2 mol}$

Tỷ lệ : $\dfrac{0,2}{1}>\dfrac{0,2}{2}$

$\text{⇒Sau pư Fe dư}$

$\text{Theo pt :}$

$n_{Fe\ pư}=1/2.n_{HCl}=1/2.0,2=0,1mol$

$⇒n_{Fe\ dư}=0,2-0,1=0,1mol$

$⇒m_{Fe\ dư}=0,1.56=5,6g$

$\text{b.Theo pt :}$

$n_{H_2}=1/2.n_{HCl}=1/2.0,2=0,1mol$

$⇒V_{H_2}=0,1.22,4=2,24l$

$\text{c.Theo pt :}$

$n_{FeCl_2}=1/2.n_{HCl}=1/2.0,2=0,1mol$

$⇒m_{FeCl_2}=0,1.127=12,7g$

$\text{Câu 2 :}$

$m_{CuO}=40\%.80=32g$

$⇒n_{CuO}=\dfrac{32}{80}=0,4mol$

$m_{Fe_2O_3}=80-32=48g$

$⇒n_{Fe_2O_3}=\dfrac{48}{160}=0,3mol$

$a.PTHH :$

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

$Fe_2O_3+3H_2O\overset{t^o}\to 2Fe+3H_2O$

$\text{b.Theo pt (1) :}$

$n_{Cu}=n_{CuO}=0,4mol$

$⇒m_{Cu}=0,4.64=25,6g$

$\text{Theo pt (2) :}$

$n_{Fe}=1/2.n_{Fe_2O_3}=1/2.0,3=0,15mol$

$⇒m_{Fe}=0,15.56=8,4g$