\(n_{N_2}=a\left(mol\right),n_{H_2}=3a\left(mol\right),n_{NH_3}=b\left(mol\right)\)

\(n_{hh}=4a+b=1\left(mol\right)\)

\(\overline{M}=\dfrac{28a+3a\cdot2+17b}{1}=6.8\cdot2=13.6\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow34a+17b=13.6\)

\(\Leftrightarrow a=0.1,b=0.6\)

\(\%V_{N_2}=10\%,\%V_{H_2}=30\%,\%V_{NH_3}=60\%\)

\(\%m_{N_2}=\dfrac{0.1\cdot28}{13.6}\cdot100\%=20.58\%\)

\(\%m_{H_2}=\dfrac{0.6}{13.6}\cdot100\%=4.41\%\)

\(\%m_{NH_3}=75.01\%\)

Gọi số mol hỗn hợp là 1 mol Gọi số mol N2 H2 NH3 là a, b , c ta có a+b+c=1 M X=6,8.2=13,6->MX=13.6g ->28a+2b+17c=13,6 lại có nH2=3nN2->3a=b ->a=0,1 b=0,3 c=0,6 ->%VN2=10% %V H2=30% V5NH3=60% ->%mN2=0,1.28/13.6=20,59% %mH2=0,3.2/13.6=4,41 ->%mNH3=75%

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:

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

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

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

help me

1, a, + 8.2=16 => CH₄

+ 8,5 . 2 = 17 => NH₃

+ 16 . 2 =32 => O₂
+ 22 . 2 = 44 => CO₂

b, + 0,138 . 29 \(\approx4\) => He

+ 1,172 . 29 \(\approx34\) => H₂S

+ 2,448 . 29 \(\approx71\Rightarrow Cl_2\)

+ 0,965 . 29 \(\approx28\) => N