Đổi:400ml=0,4l

Gọi x;2y là số mol Fe,Al

Theo gt:\(m_{hhKL}\)=\(m_{Fe}+m_{Al}\)=56x+27y.2

=56x+54y=11(1)

Ta có PTHH:

Fe+\(H_2SO_4\)->\(FeSO_4\)+\(H_2\)(1)

x..........x................x.................(mol)

4Al+6\(H_2SO_4\)->2\(Al_2(SO_4)_3\)+3\(H_2\)(2)

2y...........3y...............y.........................(mol)

Ta có:\(C_{MddH_2SO_4}\)=1M

=>\(n_{H_2SO_4}\)=1.0,4=0,4mol

Theo PTHH(1);(2):

\(n_{H_2SO_4}\)=x+3y=0,4(2)

Từ (1);(2)=>\(\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}m_{Al}=54y=54.0,1=5,4\left(g\right)\\m_{Fe}=56x=56.0,1=5,6\left(g\right)\end{matrix}\right.\)

Theo PTHH(1);(2):\(n_{FeSO_4}\)=x=0,1(mol)

\(n_{Al_2\left(SO_4\right)_3}\)=y=0,1(mol)

Vậy \(C_{M\left(FeSO_4\right)}\)=0,1:0,4=0,25M

\(C_{MAl_2\left(SO_4\right)_3}\)=0,1:0,4=0,25M

\(n_{H_2SO_4}=0,4.1=0,4\left(mol\right)\)

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

Pt: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (1)

x \(\rightarrow\dfrac{3x}{2}\) \(\rightarrow0,1mol\)

Pt: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) (2)

y \(\rightarrow y\) \(\rightarrow0,1mol\)

(1)(2) \(\Rightarrow\left\{{}\begin{matrix}1,5x+y=0,4\\27x+56y=11\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

\(m_{Al}=0,2.27=5,4\left(g\right)\)

\(m_{Fe}=0,1.56=5,6\left(g\right)\)

\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,1}{0,4}=0,25M\)

\(C_{M_{FeSO_4}}=\dfrac{0,1}{0,4}=0,25M\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Ag}=y\left(mol\right)\end{matrix}\right.\Rightarrow27x+108y=4,2\left(1\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,1                           0,05           0,15

\(\Rightarrow m_{Al}=0,1\cdot27=2,7g\)

\(\Rightarrow m_{Ag}=4,2-2,7=1,5g\)

a)\(\%m_{Al}=\dfrac{2,7}{4,2}\cdot100\%=64,28\%\)

\(\%m_{Ag}=100\%-64,28\%=35,72\%\)

b)\(m_{muối}=0,05\cdot342=17,1g\)

\(n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\)

Fe + H2SO4 → FeSO4 + H2 

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

Gọi x,y lần lượt là số mol Fe, Al

\(\left\{{}\begin{matrix}56x+27y=11\\x+\dfrac{3}{2}y=0,4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

=>\(\%m_{Fe}=\dfrac{0,1.56}{11}.100=50,91\%\)

=> %m _Al = 100 - 50,91 =49,09 %

b)Theo PT:  \(n_{H_2}=n_{H_2SO_4}=0,4\left(mol\right)\)

=> \(V_{H_2}=0,4.22,4=8,96\left(l\right)\)

c) \(CM_{FeSO_4}=\dfrac{0,1}{0,2}=0,5M\)

\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{\dfrac{0,2}{2}}{0,2}=0,5M\)

Gọi số mol Al và Fe lần lượt là a;b

$\Rightarrow 27a+56b=8,3$

Bảo toàn e ta có: $3a+3b=0,6$

Giải hệ ta được $a=b=0,1$

$\Rightarrow m_{Al}=2,7(g);m_{Fe}=5,6(g)$

\(n_{Al}=a\left(mol\right),n_{Fe}=b\left(mol\right)\)

\(m=27a+56b=8.3\left(g\right)\left(1\right)\)

\(n_{SO_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(BTe:\)

\(3a+3b=0.3\cdot2=0.6\left(2\right)\)

\(\left(1\right),\left(2\right):a=b=0.1\)

\(m_{Al}=0.1\cdot27=2.7\left(g\right)\)

\(m_{Fe}=5.6\left(g\right)\)

Fe + 2HCl----->FeCl2 +H2

x-------2x---------x--------x

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

y-------3y-----------y---------1,5y

Ta có

n\(_{H2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)

Theo bài ra ta có hệ pt

\(\left\{{}\begin{matrix}56x+27y=11\\x+1,5y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

%m\(_{Fe}=\frac{0,1.56}{11}.100\%=50,91\%\)

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

Chúc bạn học tốt

Gọi n_Na = a, n_Al=b trong 13,25g hh

=> 23a + 27b = 13,25 (l)

2Na + H₂SO₄ \(\rightarrow\) Na₂SO₄ + H₂

a --------------------------------> 0,5a (mol)

2Al + 3H₂SO₄ \(\rightarrow\) Al₂(SO₄)₃ + 3H₂

b ------------------------------------> 1,5b (mol)

=> 0,5a + 1,5b = \(\frac{9,52}{22,4}\) = 0,425 (ll)

Từ (l) và (ll) => \(\begin{cases}a=0,4\\b=0,15\end{cases}\)

a) m_Na = 0,4 . 23 = 9,2 (g)

m_Al = 4,05 (g)

b) %^mNa = \(\frac{9,2}{13,25}\) . 100% = 69,43%

%^mAl= 30,57%

a, Ta có: 27n_Al + 56n_Fe = 0,83 (1)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow n_{Al}=n_{Fe}=0,01\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,01.27}{0,83}.100\%\approx32,53\%\\\%m_{Fe}\approx67,47\%\end{matrix}\right.\)

b, n_H2SO4 = n_H2 = 0,025 (mol)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,025.98}{20\%}=12,25\left(g\right)\)