Gọi x,y lần lượt là số mol của Al,Fe

n_H2 = \(\frac{2,8}{22,4}=0,125\) mol

Pt: 2Al + 6HCl --> 2AlCl₃ + 3H₂

......x.........................x............1,5x

......Fe + 2HCl --> FeCl₂ + H₂

.......y......................y.............y

Ta có hệ pt: \(\left\{\begin{matrix} 1,5x + y = 0,125 & & \\ 133,5x + 12y = 14,025 & & \end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x = 0,03 & & \\ y = 0,08 & & \end{matrix}\right.\)

m_Al = 0,03 . 27 = 0,81 (g)

m_Fe = 0,08 . 56 = 4,48 (g)

m_hh = m_Al + m_Fe = 0,81 + 4,48 = 5,29 (g)

% m_Al = \(\frac{0,81}{5,29}\) . 100% = 15,31%

% m_Fe = \(\frac{4,48}{5,29}\) . 100% = 84,69%

Đáp án B

nH₂ = 0,4 => nHCl = 0,8 = nCl

BTKL => mA = m_muối – mCl = 39,6 – 0,8.35,5 = 11,2g

Cách 1: Dùng CT và BT điện tích
\(nNO=0,3mol\)
Mà \(nNO_3\) - trong muối \(=3nNO-=0,9mol\)
Muối \(=mKL+mNO_3-\rightarrow mKL=68,25-0,9.62=12,45\)
nFe = x mol , nAl = y mol
Muối bao gồm Fe₃+ , Al₃+ , NO₃-
BTĐT : \(3x+3y=0,9\rightarrow x+y=0,3\)
\(56x+27y=12,45\)
Giải ra \(x=0,15;y=0,15\)
\(mFe=0,15.56=8,4g\)
\(mAl=4,05g\)

Bài này dễ ợt đi

Fe + 4HNO₃ -> Fe(NO₃)₃ + NO + 2H₂O (1)

Al + 4HNO₃ -> Al(NO₃)₃ + NO + 2H₂O (2)

n_NO=0,3(mol)

Đặt n_Fe=n_Fe(NO3)3=n_NO(1)=a

n_Al=n_Al(NO3)3=n_NO(2)=b

Ta có:

\(\left\{{}\begin{matrix}242a+213b=68,25\\a+b=0,3\end{matrix}\right.\)

=>a=b=0,15

m_Fe=56.0,15=8,4(g)

m_Al=27.0,15=4,05(g)

Cách 2:

a)

Gọi số mol Al, Fe là a, b

=> 27a + 56b = 8,3

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

            a----->3a------->a------>1,5a

            Fe + 2HCl --> FeCl₂ + H₂

            b------>2b------>b----->b

=> \(1,5a+b=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

=> a = 0,1; b = 0,1

=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,1.27}{8,3}.100\%=32,53\%\\\%Fe=\dfrac{0,1.56}{8,3}.100\%=67,47\%\end{matrix}\right.\)

b) 

\(\left\{{}\begin{matrix}m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\m_{FeCl_2}=0,1.127=12,7\left(g\right)\end{matrix}\right.\)

=> m_muối = 13,35 + 12,7 = 26,05(g)

c) 

n_HCl = 3a + 2b = 0,5(mol)

=> \(V_{ddHCl\left(PTHH\right)}=\dfrac{0,5}{2}=0,25\left(l\right)\)

=> V_{dd HCl(thực tế)} = \(\dfrac{0,25.110}{100}=0,275\left(l\right)\)

d)

PTHH: 2FeCl₂ + Cl₂ --> 2FeCl₃

              0,1----------------->0,1

=> \(\left\{{}\begin{matrix}m_{FeCl_3}=0,1.162,5=16,25\left(g\right)\\m_{AlCl_3}=13,35\left(g\right)\end{matrix}\right.\)

=> m_muối = 16,25 + 13,35 = 29,6(g)

a) \(n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

          0,05<-----------0,05---->0,075

=> \(\%Al=\dfrac{0,05.27}{14,15}.100\%=9,54\%\)

=> \(\%Cu=\dfrac{14,15-0,05.27}{14,15}.100\%=90,46\%\)

b) \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)

c) \(n_{Cu}=\dfrac{14,15-0,05.27}{64}=0,2\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

          0,05->0,0375

           2Cu + O₂ --t^o--> 2CuO 

            0,2-->0,1

=> \(V_{O_2}=\left(0,1+0,0375\right).22,4=3,08\left(l\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ m_{AlCl_3}=6,675\left(mol\right)\\ n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,05\left(mol\right)\\ \Rightarrow m_A=0,05.27=1,35\left(g\right);m_{Cu}=14,15-1,35=12,8\left(g\right)\\ \%m_{Cu}=\dfrac{12,8}{14,15}.100\approx90,459\%\\ \Rightarrow\%m_{Al}\approx9,541\%\\ b,n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ \Rightarrow V=V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ 2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{O_2}=\dfrac{3}{4}.n_{Al}+\dfrac{1}{2}.n_{Cu}=\dfrac{3}{4}.0,05+\dfrac{1}{2}.0,2=0,0875\left(mol\right)\)

\(\Rightarrow V_{O_2\left(đktc\right)}=0,0875.22,4=1,96\left(l\right)\)