mn ơi! tôi viết lộn CO thành CO2 ở chỗ nung nhé

2H2 +O2 -t^o-> 2H2O (1)

2CO +O2 -t^o-> 2CO2(2)

2C4H10 +13O2 -t^o-> 8CO2 +10H2O (3)

nA=0,8(mol)

nO2=3,4(mol)

nCO2\(\approx2,16\left(mol\right)\)

giả sử nH2=x(mol)

nCO=y(mol)

nC4H10=z(mol)

=> x+y+z=0,8(mol)(I)

lại có :

0,5x +0,5y+6,5z=3,4 (II)

y+4z=2,16 (III)

từ (I,II,III)

=>\(\left\{{}\begin{matrix}x=0,14\left(mol\right)\\y=0,16\left(mol\right)\\z=0,5\left(mol\right)\end{matrix}\right.\)

=> VC4H10=11,2(l)

=> %VC4H10=62,5(%)

theo (1) : nH2O(1)=nH2=0,14(mol)

theo (3) : nH2O(3)=5nC4H10=1,5(mol)

=>\(\Sigma nH2O=1,64\left(mol\right)\)

=>a=29,52(g)

TN1: (n_Mg;n_Al;n_Cu) = (a;b;c)

=> 24a + 27b + 64c = 14,2

PTHH: Mg + 2HCl --> MgCl₂ + H₂

______a---------------------------->a

2Al + 6HCl --> 2AlCl₃ + 3H₂

b-------------------------->1,5b

=> a + 1,5b = \(\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

TN2: (n_Mg;n_Al;n_Cu) = (2a;2b;2c)

PTHH: 2Mg + O₂ --t^o--> 2MgO

______2a--->a

4Al + 3O₂ --t^o--> 2Al₂O₃

2b--->1,5b

2Cu + O₂ --t^o--> 2CuO

2c--->c

=> a + 1,5b + c = \(\dfrac{11,2}{22,4}=0,5\)

=> a=0,1 (mol); b = 0,2 (mol); c = 0,1(mol
=> \(\%Mg=\dfrac{0,1.24}{14,2}.100\%=16,9\%\)

\(m_A=m_B=\frac{44,8}{2}=22,4\left(g\right)\) (1)

Ta có: \(M_A-M_B=8\Rightarrow M_A=M_B+8\left(g\right)\) (2)

\(n_A=\frac{22,4}{M_A}\left(mol\right)\)

\(n_B=\frac{22,4}{M_B}\left(mol\right)\)

Từ (1)(2) ⇒ \(n_A< n_B\)

Ta có: \(n_B-n_A=0,05\)

\(\Leftrightarrow\frac{22,4}{M_B}-\frac{22,4}{M_A}=0,05\)

\(\Rightarrow22,4M_A-22,4M_B=0,05M_AM_B\)

\(\Leftrightarrow22,4\left(M_A-M_B\right)=0,05\left(M_B+8\right)M_B\)

\(\Leftrightarrow22,4\times8=0,05M_B^2+0,4M_B\)

\(\Leftrightarrow179,2=0,05M_B^2+0,4M_B\)

\(\Leftrightarrow0,05M_B^2+0,4M_B-179,2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}M_B=56\\M_B=-64\left(loại\right)\end{matrix}\right.\)

Vậy B là Fe

\(M_A=56+8=64\left(g\right)\) ⇒ A là Cu

2,24 mak bn