Đáp án:

`a) %m_{Fe} \approx63,3% ; %m_{Zn}= 36,7%`

`b) V_{H_2}=6,72(l)`

`c) m=19,2(g)`

Giải thích các bước giải:

 `a)`

`PT : Fe + 2HCl -> FeCl_2 + H_2`

          `Zn + 2HCl -> ZnCl_2 + H_2`

Đổi `300(ml)=0,3(l)`

`n_{HCl}=C_MxxV=2xx0,3=0,6(mol)`

Gọi `a : n_{Fe} ; b : n_{Zn} ( a,b>0)`

`=> 56a+65b=17,7(1)`

Ta có : `2a+2b=n_{HCl}=0,6(2)`

Từ `(1)` và `(2)=>a=0,2 ; b=0,1`

`=>%m_{Fe}={0,2xx56xx100%}/{17,7}\approx63,3%`

`=>%m_{Zn}=100%-63,3%=36,7%`

`b)`

Theo pt : `n_{H_2}=1/2xx``n_{HCl}=1/2xx``0,6=0,3(mol)`

`=>V_{H_2}=nxx22,4=0,3xx22,4=6,72(l)`

`c)`

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

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

Theo pt : `n_{H_2O}=n_{H_2}=0,3(mol)`

BTKL : `m_{hh}+m_{H_2}=m_{cr}+m_{H_2O`

`<=>24+(0,3xx2)=m+(0,3xx18)`

`=>m=19,2(g)`

$\text{#Cyrus}$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{HCl}} = 0,3 \times 2 = 0,6mol\\
hh:Fe(a\,mol);Zn(b\,mol)\\
\left\{ \begin{array}{l}
2a + 2b = 0,6\\
56a + 65b = 17,7
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,1\\
{m_{Fe}} = 0,2 \times 56 = 11,2g\\
\% Fe = \dfrac{{11,2}}{{17,7}} \times 100\%  = 63,3\% \\
\% Zn = 100 - 63,3 = 36,7\% \\
b)\\
{n_{{H_2}}} = \dfrac{{{n_{HCl}}}}{2} = 0,3mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
c)\\
CuO + {H_2} \to Cu + {H_2}O\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{n_{{H_2}O}} = {n_{{H_2}}} = 0,3mol\\
BTKL:\\
{m_{hh}} + {m_{{H_2}}} = {m_{cr}} + {m_{{H_2}O}}\\
 \Rightarrow m = 24 + 0,3 \times 2 - 0,3 \times 18 = 19,2g
\end{array}\)