Sửa đề : 13.9 (g) 

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

\(\Rightarrow m=27a+56b=13.9\left(1\right)\)

\(n_{H_2}=\dfrac{7.84}{22.4}=0.35\left(mol\right)\)

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

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{H_2}=1.5a+b=0.35\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.1,b=0.2\)

\(\%Al=\dfrac{0.1\cdot27}{13.9}\cdot100\%=19.42\%\)

\(\%Fe=100-19.42=80.58\%\)

Câu 2 : 

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

\(m=64a+27b=11.8\left(g\right)\left(1\right)\)

\(BTKL:m_{O_2}=18.2-11.8=6.4\left(g\right)\)

\(n_{O_2}=\dfrac{6.4}{32}=0.2\left(mol\right)\)

\(2Cu+O_2\underrightarrow{^{^{t^0}}}2CuO\)

\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)

\(n_{O_2}=0.5a+0.75b=0.2\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.1,b=0.2\)

\(\%Cu=\dfrac{0.1\cdot64}{11.8}\cdot100\%=54.23\%\)

Câu 1 : 

\(n_{HCl}=2n_{H_2}=2\cdot\dfrac{11.2}{22.4}=1\left(mol\right)\)

\(BTKL:\)

\(m_{Muối}=12.725+1\cdot36.5-0.5\cdot2=48.225\left(g\right)\)

Gọi số mol Zn, Mg, Al trong mỗi phần là x, y, z (mol)

=> 65x + 24y + 27z = 7,15 (1)

- P1: 

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

             x--->2x------->x------>x

            Mg + 2HCl --> MgCl₂ + H₂

            y----->2y------->y---->y

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

             z---->3z-------->z------>1,5z

Theo PTHH: n_HCl = 2.n_H2 = 0,5 (mol)

a = m_muối = m_{kim loại} + m_Cl = 7,15 + 0,5.35,5 = 24,9 (g)

Có: x + y + 1,5z = 0,25

- P2

PTHH: 2Zn + O₂ --t^o--> 2ZnO

             x---->0,5x

            2Mg + O₂ --t^o--> 2MgO

              y--->0,5y

            4Al + 3O₂ --t^o--> 2Al₂O₃

             z--->0,75z

=> \(n_{O_2}=\dfrac{1}{2}\left(x+y+1,5z\right)=\) 0,125 (mol)

b = m_oxit = m_{kim loại} + m_O2 = 7,15 + 0,125.32 = 11,15 (g)

\(m_{H_2SO_4}=\dfrac{100\cdot9,8\%}{100\%}=9,8g\Rightarrow n_{H_2SO_4}=0,1mol\)

\(n_{H_2}=\dfrac{1,176}{22,4}=0,0525mol\)

Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)\(\Rightarrow24x+27y=1,305\left(1\right)\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

\(\Rightarrow x+\dfrac{3}{2}y=n_{H_2SO_4}=0,1\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=-0,0825\\y=0,122\end{matrix}\right.\)

Số âm=???

a) Đặt \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow27a+56b=11\)  (1)

Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Bảo toàn electron: \(3a+2b=0,4\cdot2=0,8\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2\cdot27}{11}\cdot100\%\approx49,09\%\\\%m_{Fe}=50,91\%\end{matrix}\right.\)

b) Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{muối}=m_{AlCl_3}+m_{FeCl_2}=0,2\cdot133,5+0,1\cdot127=39,4\left(g\right)\)

c) Bảo toàn electron: \(3\cdot0,2+3\cdot0,1=2n_{SO_2}\)

\(\Rightarrow n_{SO_2}=0,45\left(mol\right)\) \(\Rightarrow V_{SO_2}=0,45\cdot22,4=10,08\left(l\right)\)

a) Gọi nAl = x, nFe = y

Có 27x + 56y = 11 (1)

Bảo toàn e

3x + 2y = 2.0,4 (2)

Từ 1 và 2 => x = 0,2, y = 0,1

\(\%mAl=\dfrac{0,2.27}{11}.100\%=49,09\%\)

\(\%mFe=100-49,09=50,91\%\)

b) BTKL: 

m muối = mkim loại + mHCl - mH2

= 11 + 0,4.2.36,5 - 0,4.2 = 39,4g

c) 

Bảo toàn e

Al => Al⁺³ + 3e                                  S⁺⁶ + 2e => S⁺⁴

0,2               0,6                                             2x          x

Fe => Fe⁺³ + 3e

0,1                  0,3

=> 2x = 0,6 + 0,3 => x = 0,45 mol

=> VSO₂ = 0,45.22,4 = 10,08 lít

2Al + 3H₂SO₄ Al₂(SO₄)₃ + 3H₂ (1)

Zn + H₂SO₄ ZnSO₄ + H₂ (2)

a;n_H2=

\(a)n_{H_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\left(1\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\left(2\right)\\ n_{Al}=a;n_{Fe}=b\\ \left\{{}\begin{matrix}1,5a+b=0,2\\27a+56b=5,5\end{matrix}\right.\\ a=0,1\\ b=0,05\\ \%_{Al}=\dfrac{0,1.27}{5,5}\cdot100=49\%\\ \%_{Fe}=100-49=51\%\\ b)n_{HCl\left(1\right)_{ }}=0,1\cdot\dfrac{6}{2}=0,3\left(mol\right)\\ n_{HCl\left(2\right)}=0,05.2=0,1\left(mol\right)\\ n_{HCl}=0,3+0,1=0,4\left(mol\right)\\ C_{M_{HCl}}=\dfrac{0.4}{0,5}=0,8M\)

nN₂O = \(\dfrac{0,616}{22,4}\)= 0,0275 mol

2N⁺⁵  +   8e  -->  N₂⁺¹

               0,22<------- 0,0275

nNO₃^- tạo muối = n_e trao đổi = 0,22 mol

=> mNO₃^- tạo muối = 0,22.62 = 13,64 gam

mMuối = mKL + mNO₃^- = 3,37 + 13,64 = 17,01 gam

Đáp án C.