\(Gọi\ n_{CO} =a(mol) ; n_{CO_2} = b(mol)\\ n_{khí} = a + b = \dfrac{15,68}{22,4} = 0,7(mol)\\ m_{khí} = 28a + 44b = 27,6(gam)\\ \Rightarrow a = 0,2 ; b = 0,5\\ \%m_{CO} = \dfrac{0,2.28}{27,6}.100\% = 20,29\%\\ \%m_{CO_2} = 100\% - 20,29\% = 79,71\%\)

a) \(n_{N_2}+n_{CO_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)

Có: \(28.n_{N_2}+44.n_{CO_2}=24,4\)

=> \(\left\{{}\begin{matrix}n_{N_2}=0,4\left(mol\right)\\n_{CO_2}=0,3\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%V_{N_2}=\dfrac{0,4}{0,7}.100\%=57,143\%\\\%V_{CO_2}=\dfrac{0,3}{0,7}.100\%=42,857\%\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}m_{N_2}=0,4.28=11,2\left(g\right)\\m_{CO_2}=0,3.44=13,2\left(g\right)\end{matrix}\right.\)

\(Đặt:n_{hh}=1\left(mol\right)\)

\(n_{NO_2}=a\left(mol\right),n_{NO}=b\left(mol\right)\)

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

\(\overline{M}=\dfrac{46a+30b}{a+b}=18.2\cdot2=36.4\)

\(\Leftrightarrow46a+30b=36.4\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.4,b=0.6\)

Tới đây tự tính tiếp nhé !!

Ta có: \(\overline{M}_{hh}=18,2\cdot2=36,4\left(đvC\right)\) 

Theo sơ đồ đường chéo: \(\dfrac{n_{NO_2}}{n_{NO}}=\dfrac{6,4}{9,6}=\dfrac{2}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{NO_2}=\dfrac{2}{5}\cdot100\%=40\%\\\%V_{NO}=60\%\end{matrix}\right.\)

Giả sử \(n_{NO_2}=2\left(mol\right)\) \(\Rightarrow n_{NO}=3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{NO_2}=\dfrac{2\cdot46}{2\cdot46+3\cdot30}\cdot100\%\approx50,55\%\\\%m_{NO}=49,45\%\end{matrix}\right.\)

ko phải e làm đâu nha a , a tham khảo cho 

a)CH4 + 2O2 →→ CO2 + 2H2O (1)

2H2 + O2 →→ 2H2O (2)

b) Đặt nCH4 = a (mol) , nH2 = b (mol)

=> mCH4 = 16a(g) , mH2 = 2b(g)

mà n(CH4 + H2) = V/22,4 = 6,72/22,4 = 0,3(mol)

=> a + b = 0,3(mol) => a = 0,3 - b

Từ PT(1) => nO2 = 2nCH4 = 2a(mol)

Từ PT(2) => nO2 = 1/2 . nH2 = 1/2 . b (mol)

=>tổng nO2 = 2a + 1/2.b (mol)

=> mO2 = n .M = 32. (2a+ 1/2.b )= 64a + 16b(g)

Theo ĐLBTKL:mCH4 + mH2 + mO2 = mCO2 + mH2O(PT1,2)

=> 16a + 2b+ 64a +16b = 11,6

=> 80a + 18b = 11,6

=> 80. (0,3 - b ) + 18b = 11,6

=> b = 0,2(mol)

=> a = 0,3 - 0,2 = 0,1(mol)

=>mCH4 = 16a = 16. 0,1 = 1,6(g) , mH2 = 2b = 2 . 0,2 = 0,4(g)

=> VCH4 = n .22,4 = 0,1 . 22,4 =2,24(l)

VH2 = n . 22,4 = 0,2 . 22,4 = 4,48(l)

tự tính ... 

c) nO2 = 2a + 1/2 .b = 2 . 0,1 + 1/2 . 0,2 = 0,3(mol)

=> VO2 = 0,3 . 22,4 =6,72(l)

mà VO2 = 20% Vkk

=> Vkk = 6,72 : 20% =33,6(l)

\(\%mCO2=\dfrac{22}{22+6,4+16}.100\approx49,55\%\\ \%mSO2=\dfrac{6,4}{22+6,4+16}.100\approx14,414\%\\ \%mO2=\dfrac{16}{22+6,4+16}.100\approx36,036\%\)

a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Ta có: \(n_{CH_4}+n_{C_2H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(1\right)\)

\(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,25\left(mol\right)\\n_{C_2H_2}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,25.22,4}{6,72}.100\%\approx83,33\%\\\%V_{C_2H_2}\approx16,67\%\end{matrix}\right.\)

Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=0,625\left(mol\right)\Rightarrow m_{O_2}=0,625.32=20\left(g\right)\)

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH : \(C+O_2\rightarrow CO_2\)

            x        x        x        (mol)

PTHH : \(S+O_2\rightarrow SO_2\)

           y          y          y       (mol)

\(\rightarrow\) x + y = 0,3 (1)

12x+32y=5,6  (2)

Từ (1),(2) \(\rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

\(\%C=\dfrac{0,2.12}{5,6}.100=42,86\%\)

\(\%S=100\%-42,86\%=57,14\%\)

\(m_C=0,2.12=2,4\left(g\right)\)

\(m_S=5,6-2,4=3,2\left(g\right)\)

nO2 = 9.6/32 = 0.3 (mol) 

nCO2 = 8.8/44 = 0.2 (mol) 

CO + 1/2O2 -to-> CO2 

0.2_____0.1______0.2 

H2 + 1/2O2 -to-> H2O 

0.4__0.3-0.1

%CO = 0.2*28/(0.2*28 + 0.4*2) * 100% = 87.5%

Chúc bạn học tốt !!!