Gọi hh khí ban đầu là X và hh khí sản phẩm là Y.

\(d_{\dfrac{X}{H_2}}=3,6\rightarrow M_X=3,6.2=7,2\left(\dfrac{g}{mol}\right)\)

Ta có:

\(\dfrac{n_{N_2}}{n_{H_2}}=\dfrac{7,2-2}{28-7,2}=\dfrac{5,2}{20,8}=\dfrac{1}{4}\) ( quy tắc đường chéo )

Gọi x là số mol N2 tham gia phản ứng.

PTHH:

               \(N_2+3H_2\xrightarrow[Fe]{t^o}2NH_3\)

Trc p/ư:   1         4 

p/ư:         x       3x             2x               (mol)

sau p/ư:  1-x     4-3x         2x  

\(\rightarrow n_Y=1-x+4-3x+2x=5-2x\left(mol\right)\)

\(d_{\dfrac{Y}{H_2}}=4,5\rightarrow M_Y=4,5.2=9\left(\dfrac{g}{mol}\right)\)

BTKL: m_X = m_Y 

\(m_Y=1.28+4.2=36\left(g\right)\)

\(\rightarrow\dfrac{36}{9}=5-2x\)

\(\rightarrow x=0,5\)

Giả sử phản ứng xảy ra hoàn toàn: 

\(\dfrac{1}{1}< \dfrac{4}{3}\) --> N2 hết, H2 dư.

Trước:

\(\%V_{N_2}=\dfrac{1.22,4}{5.22,4}.100\%=20\%\)

\(\%V_{H_2}=100\%-20\%=80\%\)

Sau:

\(\%V_{NH_3}=\dfrac{1.22,4}{4.22,4}.100\%=25\%\)

\(\%V_{N_2}=\dfrac{\left(1-0,5\right).22,4}{4.22,4}.100\%=12,5\left(\%\right)\)

\(\%V_{H_2}=\dfrac{\left(4-1,5\right).22,4}{4.22,4}.100\%=62,5\%\)

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

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

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

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

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

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

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

\(N_2+3H_2⇌2NH_3\)

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

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

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

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

\(\Leftrightarrow a=0.1\)

\(\%N_2=12.5\%\)

\(\%H_2=62.5\%\)

\(\%NH_3=25\%\)

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

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 : 

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

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

\(a.\)

\(GS:\)

\(n_{hh}=1\left(mol\right)\)

\(Đặt:n_{N_2}=a\left(mol\right),n_{CO_2}=b\left(mol\right)\)

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

\(m_A=28a+44b=18\cdot2=36\left(2\right)\)

\(\left(1\right),\left(2\right):\)

\(a=b=0.5\)

\(\%m_{N_2}=\dfrac{0.5\cdot28}{0.5\cdot28+0.5\cdot44}\cdot100\%=38.89\%\)

\(\%m_{CO_2}=61.11\%\)

\(b.\)

\(\dfrac{n_{N_2}}{n_{CO_2}}=\dfrac{0.5}{0.5}=\dfrac{1}{1}\)

\(n_{N_2}=n_{CO_2}=\dfrac{1}{2}\cdot n_A=\dfrac{0.2}{2}=0.1\left(mol\right)\)

\(Đặt:n_{CO_2}=x\left(mol\right)\)

\(\overline{M}=\dfrac{0.1\cdot28+0.1\cdot44+44x}{0.2+x}=20\cdot2=40\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow x=0.2\)

\(m_{CO_2\left(cầnthêm\right)}=0.2\cdot44=8.8\left(g\right)\)

3,6 hay 36 :) ??