\(n_A=1\left(mol\right)\)

\(n_{N_2}=a\left(mol\right),n_{H_2}=b\left(mol\right)\)

\(\Leftrightarrow a+b=1\left(1\right)\)

\(m_A=28a+2b=7.2\left(g\right)\left(2\right)\)

\(\left(1\right)\left(2\right):a=0.2,b=0.8\)

\(\%N_2=20\%,\%H_2=80\%\)

\(n_{N_2\left(pư\right)}=a\left(mol\right)\)

\(N_2+3H_2⇌2NH_3\)

\(0.2......0.8\)

\(a.......3a.........2a\)

\(0.2-a.0.8-3a....2a\)

\(M_B=\dfrac{\left(0.2-a\right)\cdot28+\left(0.8-3a\right)\cdot2+2a\cdot17}{0.2-a+0.8-3a+2a}=9\)

\(\Leftrightarrow a=0.1\)

\(\%N_2=12.5\%\)

\(\%H_2=62.5\%\)

\(\%NH_3=25\%\)

\(H\%=\dfrac{0.1}{0.2}\cdot100\%=50\%\)

Coi $n_A = 1(mol)$
Gọi $n_{N_2} = a ; n_{H_2} = b$

$M_A = 3,6.2 = 7,2$

Ta có: 

$a + b = 1$
$28a + 2b = 7,2(a + b)$
Suy ra a = 0,2;  b = 0,8

Vậy $\%V_{N_2} = \dfrac{0,2}{1}.100\% = 20% ; \%V_{H_2} = 80\%$

Gọi hiệu suất là a

$N_2 + 3H_2 \xrightarrow{t^o,xt} 2NH_3$

Ta thấy : $n_{N_2} : 1 < n_{H_2} : 3$ nên hiệu suất tính theo $N_2$

$n_{N_2\ pư} = 0,2a(mol)$

Theo PTHH : 

$n_{H_2\ pư} = 0,6a(mol) ; n_{NH_3} = 0,4a(mol)$

$m_B = m_A = 7,2(gam)$

$\Rightarrow n_B = \dfrac{7,2}{4,5.2} = 0,8$

Khí B gồm : 

$N_2 : 0,2 - 0,2a(mol)$
$H_2 : 0,8 - 0,6a(mol)$
$NH_3 : 0,4a(mol)$

Suy ra : 0,2 - 0,2a + 0,8 - 0,6a + 0,4a = 0,8 

Suy ra a = 0,5 = 50%

Vậy B gồm : 

$N_2 : 0,1(mol)$
$H_2 : 0,5(mol)$
$NH_3 : 0,2(mol)$
$\%V_{N_2} = \dfrac{0,1}{0,8}.100\% = 12,5\%$
$\%V_{H_2} = \dfrac{0,5}{0,8}.100\% = 62,5\%$

$\%V_{NH_3} = 25\%$

Đáp án D

$C_2H_2+2H_2\xrightarrow{t^o,Ni}C_2H_6(1)$

$C_2H_2+2AgNO_3+2NH_3\to C_2Ag_2\downarrow+NH_4NO_3(2)$

$C_2H_2+2Br_2\to C_2H_2Br_4(3)$

$2C_2H_6+7O_2\xrightarrow{t^o}4CO_2+6H_2O$

Theo PT: 

$n_{C_2H_2(2)}=n_{C_2Ag_2}=\frac{24}{240}=0,1(mol)$

$n_{C_2H_2(3)}=\frac{1}{2}n_{Br_2}=\frac{1}{2}.\frac{40}{160}=0,125(mol)$

$n_{C_2H_6}=\frac{1}{3}n_{H_2O}=\frac{1}{3}.\frac{11,7}{18}=\frac{13}{60}(mol)$

Sau pứ (1):

$n_{C_2H_2(dư)}=0,125+0,1=0,225(mol)$

Theo PT(1): $n_{C_2H_2(pứ)}=\frac{13}{60}(mol);n_{H_2}=\frac{13}{30}(mol)$

$\to a=n_{C_2H_2}+n_{H_2}=\frac{13}{60}+\frac{13}{30}+0,225=\frac{7}{8}(mol)$

a) \(N_2+3H_2-^{t^o}\rightarrow2NH_3\)

    3..........8

Lập tỉ lệ \(\dfrac{3}{1}>\dfrac{8}{3}\) => Tính theo số mol H2

\(n_{H_2\left(pứ\right)}=8.60\%=4,8\left(mol\right)\)

=> \(n_{NH_3}=\dfrac{4,8.2}{3}=3,2\left(mol\right)\)

=> \(V_{NH_3}=3,2.22,4=71,68\left(l\right)\)

b) Hỗn hợp A gồm N2 dư, H2 dư, NH3

\(n_{N_2\left(dư\right)}=3-\dfrac{4,8}{3}=1,4\left(mol\right)\)

\(n_{H_2\left(dư\right)}=8-4,8=3,2\left(mol\right)\)

\(M_A=\dfrac{3,2.2+1,4.28+3,2.17}{3,2+1,4+3,2}=12,82\)

=>dA/H2 \(=\dfrac{12,82}{2}=6,41\)

Chọn C

Theo bài ra có khối lượng chất rắn giảm chính bằng khối lượng O bị lấy ra khỏi oxit.

Bảo toàn electron: