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

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

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

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

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

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

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

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

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

\(\%Al=49.09\%\)

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

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

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

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

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

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

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

\(\%Al=24.33\%\)

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
$\%m_{Fe} = \dfrac{0,3.56}{22,2}.100\% =75,68\%$

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

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

3Cu  + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O     (2)

\(n_{NO}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

Đặt x và y là số mol Fe và Cu trong hỗn hợp ta có hệ pt:

56x + 64y = 3,04

\(x+\dfrac{2y}{3}=0,04\)

→ x= 0,02 , y= 0,03

\(\%m_{Fe}=\dfrac{0,02.56}{3,04}.100=36,8\%\)

%m Cu= 100 – 36,8= 63,2%.

a)

Gọi $n_{Mg} = a(mol) ; n_{Zn} = b(mol)$

Suy ra: $24a + 65b = 24,3(1)$
$n_{SO_2} = 0,5(mol)$

Bảo toàn electron : $2n_{Mg} + 2n_{Zn} = 2n_{SO_2}$
$\Rightarrow a + b = 0,5(2)$

Từ (1)(2) suy ra a = 0,2 ; b = 0,3

$m_{Mg} = 0,2.24 = 4,8(gam)$
$m_{Zn} = 0,3.65 = 19,5(gam)$
b)

Ta có :

$n_{H_2SO_4} = 2n_{SO_2} = 1(mol)$
$m_{H_2SO_4\ pư} = 1.98 = 98(gam)$

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

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

\(n_{SO_2}=\dfrac{V_{SO_2}}{22,4}=\dfrac{2,24}{22,4}=0,1mol\)

Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\\n_{Cu}=y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=24x\\m_{Cu}=64y\end{matrix}\right.\)

\(Mg+2H_2SO_4\rightarrow MgSO_4+SO_2+2H_2O\) 

 x           2x                               x                  ( mol )

\(Cu+2H_2SO_4\rightarrow CuSO_4+SO_2+2H_2O\)

 y            2x                              y                   ( mol )

Ta có:

\(\left\{{}\begin{matrix}24x+64y=4,4\\x+y=0,1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)

\(\Rightarrow m_{Mg}=0,05.24=1,2g\)

\(\Rightarrow m_{Cu}=0,05.64=3,2\)

\(m_{H_2SO_4}=n_{H_2SO_4}.M_{H_2SO_4}=\left(2.0,05+2.0,05\right).98=0,2.98=19,6g\)

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!

\(n_{SO_2}=\dfrac{2,24}{22,4}=0,1mol\)

Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Cu}=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}24x+64y=4,4\\BTe:2x+2y=2n_{SO_2}=2\cdot0,1=0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)

\(m_{Mg}=0,05\cdot24=1,2g\)

\(m_{Cu}=0,05\cdot64=3,2g\)

\(n_{H_2SO_4}=2n_{Mg}+2n_{Cu}=2\cdot0,05+2\cdot0,05=0,2mol\)

\(m_{H_2SO_4}=0,2\cdot98=19,6g\)

Đặt nMg=x(mol),nCu=y(mol)

Bảo toàn e : 2nMg+2nCu=2nSO2=0,2(mol)

Từ  giải được:

mMg=0,05.24=1,2(g)

mCu=0,05.64=3,2(g)

nH2SO4=2nSO2=0,2(mol)

mH2SO4=0,2.98=19,6(g)

Đặt nMg=x(mol);nCu=y(mol)

→24x+64y=4,4(1)

nSO2=\(\dfrac{2,24}{22,4}\)=0,1(mol)

Bảo toàn e: x+y=nSO2=0,1(2)

Từ (1)(2)→x=y=0,05(mol)

=>mMg=0,05.24=1,2(g)

mCu=0,05.64=3,2(g)

b)

Mg+2H2SO4→MgSO4+SO2+3H2O

Cu+2H2SO4→CuSO4+SO2+2H2O

Tho PT: ∑nH2SO4=2∑nSO2=0,2(mol)

→m_axit=0,2.98=19,6(g)