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

\(m_{hh}=56a+27b=11\left(g\right)\left(1\right)\)

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

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

\(n_{H_2}=a+1.5b=0.4\left(mol\right)\left(2\right)\)

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

\(\%Fe=\dfrac{0.1\cdot56}{11}\cdot100\%=50.91\%\)

\(\%Al=49.09\%\)

Gọi $n_{Fe} = a(mol) ; n_{Mg} = b(mol) \Rightarrow 56a + 24b = 9,2(1)$
$n_{NO} = 0,2(mol)$
Bảo toàn electron : 

$3a + 2b = 0,2.3(2)$
Từ (1)(2) suy ra a = 0,1 ; b = 0,15

$m_{Fe} = 0,1.56 = 5,6(gam)$
$m_{Mg} = 0,15.24 = 3,6(gam)$

Viết pt đc kh ạ

\(Đặt:\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}24x+27y=10,2\\x+1,5y=0,5\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,2.24}{10,2}.100=47,06\%\\ \%m_{Al}=52,94\%\\ n_{HCl}=2n_{Mg}+3n_{Al}=0,2.2+0,3.2=1\left(mol\right)\\ \Rightarrow V_{HCl}=\dfrac{1}{2}=0,5\left(l\right)\)

\(c,n_{Al}=x(mol);n_{Mg}=y(mol)\\ \Rightarrow 27x+24y=10,2(1)\\ n_{H_2}=\dfrac{11,2}{22,4}=0,5(mol)\\ PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ Mg+2HCl\to MgCl_2+H_2\\ \Rightarrow 1,5x+y=0,5(2)\\ (1)(2)\Rightarrow x=y=0,2(mol)\\ \Rightarrow \%_{Al}=\dfrac{0,2.27}{10,2}.100\%=52,94\%\\ \%_{Mg}=100\%-52,94\%=47,06\%\\ d,\Sigma n_{HCl}=3x+2y=1(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{1}{2}=0,5(l)\)

a) Gọi số mol Fe, Cr là a, b (mol)

=> 56a + 52b = 10,8 (1)

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

PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂

             a---->a------------------->a

             Cr + H₂SO₄ --> CrSO₄ + H₂

              b--->b------------------->b

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

(1)(2) => a = 0,1 (mol); b = 0,1 (mol)

=> \(\left\{{}\begin{matrix}\%Fe=\dfrac{0,1.56}{10,8}.100\%=51,85\%\\\%Cr=\dfrac{0,1.52}{10,8}.100\%=48,15\%\end{matrix}\right.\)

b) \(n_{H_2SO_4}=a+b=0,2\left(mol\right)\)

=> \(m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)

cam on ah

a)

Gọi $n_{Mg} = a ; n_{Al} = b \Rightarrow 24a + 27b = 9,9(1)$

$n_{NO} = 0,25(mol)$
Bảo toàn e: $2a + 3b = 0,25.3(2)$
Từ (1)(2) suy ra a = 0,525 ; b = -0,1<0

$\to$ Sai đề

3Mg + 8HNO3 → 3Mg(NO3)2 + 2NO + 4H2O

Al + 4HNO3 → Al(NO3)3 + NO + 2H2O

Gọi x, y lần lượt là số mol Mg, Al

\(\left\{{}\begin{matrix}24x+27y=9,9\\\dfrac{2}{3}x+y=\dfrac{5,6}{22,4}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=0,525\\y=-0,1\end{matrix}\right.\)

Bạn xem lại đề nhé!

Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Cu}=b\left(mol\right)\end{matrix}\right.\)

\(Fe+4HNO_{3l}\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\)

\(3Cu+8HNO_{3l}\rightarrow3Cu\left(NO_3\right)_2+2NO+4H_2O\)

\(n_{NO}=\dfrac{6,72}{22,4}=0,3mol\)

\(\Rightarrow\left\{{}\begin{matrix}56a+64b=24,8\\BTe:3a+2b=3\cdot0,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,3\end{matrix}\right.\)

\(\%m_{Fe}=\dfrac{0,1\cdot56}{24,8}\cdot100\%=22,58\%\)

\(\%m_{Cu}=100\%-22,58\%=77,42\%\)

cam on ah

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

\(m_{hh}=56a+27b=22.2\left(g\right)\left(1\right)\)

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

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

\(n_{H_2}=a+1.5b=0.6\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.3,b=0.2\)

\(\%Fe=\dfrac{0.3\cdot56}{22.2}\cdot100\%=75.67\%\)

\(\%Al=24.33\%\)

Gọi $n_{Fe} = a ; n_{Al} = b$

$\Rightarrow 56a + 27b = 22,2(1)$
$Fe + H_2SO_4 \to FeSO_4 + H_2$
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$

Theo PTHH : 

$n_{H_2} = a + 1,5b = 0,6(2)$
Từ (1)(2) suy ra a = 0,3;  b = 0,2
$\%m_{Fe} = \dfrac{0,3.56}{22,2}.100\% =75,68\%$

$\%m_{Al} = 24,32\%$