Gọi số mol Fe, Zn là a,b

=> 56a + 65b = 12,1 (1)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

______a---->2a

Zn + 2HCl --> ZnCl₂ + H₂

_b---->2b

=> 2a + 2b = 0,4 (2)

(1)(2) => \(\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

=> \(\%Fe=\dfrac{56.0,1}{12,1}.100\%=46,28\%\)

Đặt \(n_{Fe}=x(mol);n_{Zn}=y(mol)\Rightarrow 56x+65y=12,1(1)\)

\(n_{HCl}=2.0,2=0,4(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow 2x+2y=0,4(2)\\ (1)(2)\Rightarrow x=y=0,1(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,1.56}{12,1}.100\%=46,28\%\)

\(n_{H_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\) . Gọi x và y lần lượt là số mol của Fe và Zn tham gia phản ứng.

a/ PTHH : Fe + 2HCl -----> FeCl₂ + H₂

               x          2x                          x

            Zn + 2HCl ------> ZnCl₂ + H₂

             y       2y                           y

Ta có hệ : \(\begin{cases}56x+65y=12,1\\x+y=0,2\end{cases}\) \(\Leftrightarrow\begin{cases}x=0,1\\y=0,1\end{cases}\)

\(\Rightarrow m_{Fe}=0,1\times56=5,6\left(g\right)\)

\(\Rightarrow\%Fe=\frac{5,6}{12,1}\times100\approx46,28\%\)

\(\Rightarrow\%Zn=100\%-46,28\%=53,72\%\)

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Đặt :

nFe = x mol

nAl = y mol

Fe + 2HCl --> FeCl2 + H2

x____________x______x

2Al + 6HCl --> 2AlCl3 + 3H2

y______________y______1.5y

mhh= 56x + 27y = 5.5 g (1)

nH2 = x + 1.5y = 0.2 (2)

Giải (1) và (2) :

x = 0.05

y= 0.1

mFe = 2.8 g

mAl = 2.7 g

%Fe = 50.91%

%Al = 49.09%

Từ PTHH ta thấy :

nHCl = 2nH2 = 0.2*2=0.4 mol

mHCl = 0.4*36.5 = 14.6 g

mddHCl = 14.6*100/14.6 = 100g

mdd sau phản ứng = 5.5 + 100 - 0.4 = 105.1 g

mFeCl2 = 6.35g

mAlCl3 = 13.35g

C%FeCl2 = 6.04%

%AlCl3 = 12.7%

Fe + 2HCl → FeCl₂ + H₂ (1)

2Al + 6HCl → 2AlCl₃ + 3H₂ (2)

\(n_{H_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

a) Gọi x,y lần lượt là số mol của Fe và Al

Ta có: \(56x+27y=5,5\) (*)

Theo Pt1: \(n_{H_2}=n_{Fe}=x\left(mol\right)\)

Theo PT2: \(n_{H_2}=\frac{3}{2}n_{Al}=1,5y\left(mol\right)\)

Ta có: \(x+1,5y=0,2\) (**)

Từ (*)(**) ta có: \(\left\{{}\begin{matrix}56x+27y=5,5\\x+1,5y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)

Vậy \(n_{Fe}=0,05\left(mol\right)\Rightarrow m_{Fe}=0,05\times56=2,8\left(g\right)\)

\(n_{Al}=0,1\left(mol\right)\Rightarrow m_{Al}=0,1\times27=2,7\left(g\right)\)

\(\%m_{Fe}=\frac{2,8}{5,5}\times100\%=50,91\%\)

\(\%m_{Al}=100\%-50,91\%=49,09\%\)

b) Theo Pt1: \(n_{HCl}=2n_{Fe}=2\times0,05=0,1\left(mol\right)\)

Theo Pt2: \(n_{HCl}=3n_{Al}=3\times0,1=0,3\left(mol\right)\)

\(\Rightarrow\Sigma n_{HCl}=0,1+0,3=0,4\left(mol\right)\)

\(\Rightarrow\Sigma m_{HCl}=0,4\times36,5=14,6\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\frac{14,6}{14,6\%}=100\left(g\right)\)

c) \(m_{H_2}=0,2\times2=0,4\left(g\right)\)

Ta có: \(m_{dd}saupứ=5,5+100-0,4=105,1\left(g\right)\)

Theo Pt1: \(n_{FeCl_2}=n_{Fe}=0,05\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=0,05\times127=6,35\left(g\right)\)

\(\Rightarrow C\%_{FeCl_2}=\frac{6,35}{105,1}\times100\%=6,04\%\)

Theo Pt2: \(n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,1\times133,5=13,35\left(g\right)\)

\(\Rightarrow C\%_{AlCl_3}=\frac{13,35}{105,1}\times100\%=12,7\%\)

Đặt \(n_{Fe}=x(mol);n_{Zn}=y(mol)\Rightarrow 56x+65y=18,6(1)\)

\(n_{HCl}=3.0,2=0,6(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow 2x+2y=0,6(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ \Rightarrow m_{Zn}=0,2.65=13(g)\)

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

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

Đặt \(n_{Fe}=x(mol);n_{Zn}=y(mol)\Rightarrow 56x+65y=9,3(1)\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow x+y=0,15(2)\\ (1)(2)\Rightarrow x=0,05(mol);y=0,1(mol)\\ \Rightarrow \%_{Zn}=\dfrac{0,1.65}{9,3}.100\%=69,89\%\)