a)

$Zn + S \xrightarrow{t^o} ZnS$

$n_{Zn} =\dfrac{9,75}{65} = 0,15 > n_S = \dfrac{3,84}{32} = 0,12$ nên Zn dư

$n_{ZnS} = n_S = 0,12(mol)$
$m_{ZnS} = 0,12.97 = 11,64(gam)$
$n_{Zn\ dư} = 0,15 - 0,12 = 0,03(mol)$
$m_{Zn\ dư} = 0,03.65 = 1,95(gam)$

b)

$Zn + 2HCl \to ZnCl_2 + H_2$
$ZnS + 2HCl \to ZnCl_2 + H_2S$
$n_{khí} = n_{H_2} + n_{H_2S} = n_{Zn\ dư} + n_{ZnS} = 0,15(mol)$
$V = 0,15.22,4 = 3,36(lít)$

\(n_{Zn}=\dfrac{9.75}{65}=0.15\left(mol\right)\)

\(n_S=\dfrac{3.84}{32}=0.12\left(mol\right)\)

\(Zn+S\underrightarrow{^{^{t^0}}}ZnS\)

Lập tỉ lệ : 

\(\dfrac{0.15}{1}>\dfrac{0.12}{1}\Rightarrow Zndư\)

\(a.\)

\(m_X=m_{ZnS}+m_{Zn\left(dư\right)}=0.12\cdot97+\left(0.15-0.12\right)\cdot65=13.59\left(g\right)\)

\(b.\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.03..................................0.03\)

\(ZnS+2HCl\rightarrow ZnCl_2+H_2S\)

\(0.12.................................0.12\)

\(V_{khí}=0.03\cdot22.4+0.12\cdot22.4=3.36\left(l\right)\)

a) Đặt \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow56a+27b=1,93\)  (1)

Ta có: \(n_S=\dfrac{1,28}{32}=0,04\left(mol\right)\)

Bảo toàn electron: \(2a+3b=0,08\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{121}{3800}\left(mol\right)\\b=\dfrac{31}{5700}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=\dfrac{121}{3800}\cdot56\approx1,78\left(g\right)\\m_{Al}\approx0,15\left(g\right)\end{matrix}\right.\)

b) Bảo toàn nguyên tố: \(\Sigma n_{H_2S}=n_{FeS}+3n_{Al_2S_3}=n_{Fe}+6n_{Al}=\dfrac{49}{760}\left(mol\right)\)

\(\Rightarrow V_{H_2S}=\dfrac{49}{760}\cdot22,4\approx1,44\left(l\right)\)

Đáp án A.

Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\)

\(n_{HCl}=0,2\cdot4=0,8mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(x\)   \(\rightarrow\)   \(3x\)            \(x\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

 \(y\)   \(\rightarrow\) \(2y\)            \(y\)

\(\Rightarrow\left\{{}\begin{matrix}27x+65y=11,9\\3x+2y=0,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

a)\(\%m_{Al}=\dfrac{0,2\cdot27}{11,9}\cdot100\%=45,38\%\)

\(\%m_{Zn}=100\%-45,38\%=54,62\%\)

b)\(\Sigma n_{H_2}=\dfrac{3}{2}x+y=\dfrac{3}{2}\cdot0,2+0,1=0,4mol\)

\(V_{H_2}=0,4\cdot22.4=8,96l\)

Qui hỗn hợp về dạng : x mol Fe ; y mol FeO ; z mol Fe₂O₃

=> m_hh = 56x + 72y + 160z = 26 (1)

Có n_HCl = 2x + 2y + 6z = 0,91 mol (2)

Và n_H2 = n_Fe = x = 0,195 mol (3)

Từ (1) ; (2) ; (3) => y = z = 0,065 mol

=> Khi nung kết tủa thì sản phẩm chỉ còn Fe₂O₃

Bảo toàn Fe : n_Fe2O3 = ½ n_{Fe(hh đầu)} = ½ . ( x + y + 2z) = 0,195 mol

=>m = 31,2g

=>B