Gọi \(\left\{{}\begin{matrix}n_{CH_4}=x\left(mol\right)\\n_{C_2H_6}=y\left(mol\right)\end{matrix}\right.\)

\(n_{H_2O}=\dfrac{14,4}{18}=0,8\left(mol\right)\)

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

x-------->2x----------->2x

\(C_2H_6+\dfrac{7}{2}O_2\underrightarrow{t^o}2CO_2+3H_2O\)

y-------->3,5y------------->3y

Có hệ phương trình: \(\left\{{}\begin{matrix}x+y=\dfrac{6,72}{22,4}=0,3\\2x+3y=0,8\end{matrix}\right.\)

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

a

\(\%V_{CH_4}=\dfrac{0,1.22,4.100\%}{6,72}=33,33\%\)

\(\%V_{C_2H_6}=\dfrac{0,2.22,4.100\%}{6,72}=66,67\%\)

b

\(V_{O_2}=\left(2x+3,5y\right).22,4=\left(2.0,1+3,5.0,2\right).22,4=20,16\left(l\right)\)

Cho dữ liệu dư 10% như thế thì phải hỏi là V khí \(O_2\) đã lấy/ đã dùng chứ "cần lấy" là theo PTHH (không cần cho "Biết ...")

\(V_{O_2.đã.lấy}=\dfrac{20,16.\left(100+10\right)\%}{100\%}=22,176\left(l\right)\)

Làm giúp em câu b với ạ, em cảm ơn

a, Ta có: \(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)=n_C\)

\(n_{H_2O}=\dfrac{10,8}{18}=0,6\left(mol\right)\Rightarrow n_H=0,6.2=1,2\left(mol\right)\)

m = m_C + m_H = 0,4.12 + 1,2.1 = 6 (g)

b, Theo ĐLBT KL, có: m + m_O2 = m_CO2 + m_H2O

⇒ m_O2 = 22,4 (g) \(\Rightarrow n_{O_2}=\dfrac{22,4}{32}=0,7\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,7.22,4=15,68\left(g\right)\)

\(\Rightarrow V_{kk}=\dfrac{15,68}{20\%}=78,4\left(g\right)\)

\(a,Đặt:n_{CH_4}=a\left(mol\right);n_{C_4H_{10}}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ 2C_4H_{10}+13O_2\rightarrow\left(t^o\right)8CO_2+10H_2O\\ \Rightarrow\left\{{}\begin{matrix}16a+58b=7,4\\22,4a+22,4.4b=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{CH_4}=0,1.16=1,6\left(g\right)\\m_{C_4H_{10}}=0,1.58=5,8\left(g\right)\end{matrix}\right.\\ b,n_{O_2}=2a+\dfrac{13}{2}b=2.0,1+6,5.0,1=0,85\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,85.22,4=19,04\left(l\right)\)

a) PTHH: \(2CO+O_2\underrightarrow{t^o}2CO_2\)  (1)

                \(4H_2+O_2\underrightarrow{t^o}2H_2O\)  (2)

b) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\\n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(1\right)}=0,1mol\\n_{O_2\left(2\right)}=0,2mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CO}=0,1\cdot28=2,8\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{CO}=\dfrac{2,8}{2,8+0,4}\cdot100\%=87,5\%\\\%m_{H_2}=12,5\%\end{matrix}\right.\)

c) PTHH: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

Theo PTHH: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)

\(\Rightarrow m_{KMnO_4}=0,6\cdot158=94,8\left(g\right)\)

C₂H₂ + H₂ C₂H₄

C₂H₂ + H₂ C₂H₆

Khi cho hỗn hợp B qua dd nước Br₂ chỉ có C₂H₄ và C₂H₂ phản ứng C₂H₂ +2Br₂ → C₂H₂Br₄

C₂H₄ + Br₂ → C₂H₄Br₂

=> khối lượng bình Br₂ tăng chính bằng khối lượng của C₂H₂ và C₂H₄

m_C2H2 + m_C2H4 = 4,1 (g)    

Hỗn hợp khí D đi ra là CH₄, C₂H₆ và H₂

CH₄ + 2O₂ → t ∘  CO₂ + 2H₂O

C₂H₆ + O₂  → t ∘  2CO₂ + 3H₂O

2H₂ + O₂  → t ∘  2H₂O

Bảo toàn nguyên tố O cho quá trình đốt cháy hh D ta có:

2n_CO2 = 2n_O2 – n_H2O => n_CO2 = ( 0,425. 2– 0,45)/2 = 0,2 (mol)

Bảo toàn khối lượng : m_{hh D} = m_CO2 + m_H2O – m_O2 = 0,2.44 + 0,45.18 – 0,425.32 = 3,3 (g)

Bảo toàn khối lương: m_A = (m_C2H2 + m_C2H4) + m_{hh D} = 4,1 + 3,3 = 7,4 (g)

\(2CO + O_2 \xrightarrow{t^o} 2CO_2(1)\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O(2)\\ n_{CO} = n_{CO_2} = \dfrac{4,48}{22,4} = 0,2(mol)\\ n_{O_2(1)} = \dfrac{n_{CO_2}}{2} = 0,1(mol)\\ \Rightarrow n_{H_2} = 2n_{O_2(2)} = 2.(\dfrac{6,72}{22,4}-0,1)=0,4(mol)\\ \%V_{CO} = \dfrac{0,2}{0,2+0,4}.100\% = 33,33\%\\ \%V_{H_2} = 100\% -33,33\% = 66,67\%\)

\(\%m_{CO} = \dfrac{0,2.28}{0,2.28+0,4.2}.100\% = 87,5\%\\ \%m_{H_2} = 100\% - 87,5\% = 12,5\%\)

nO (trong CO2) = 2 . nCO2 = 2 . 26,4/44 = 1,2 (mol)

nO (trong H2O) = nH2O = 13,5/18 = 0,75 (mol)

nO (trong O2) = 1,2 + 0,75 = 1,95 (mol)

nO2 = 1,95/2 = 0,975 (mol)

VO2 = 0,975 . 22,4 = 21,84 (l)

Vkk = 21,84 . 5 = 109,2 (l)

nO (trong CO2) = 2 . nCO2 = 2 . 26,4/44 = 1,2 (mol)

nO (trong H2O) = nH2O = 13,5/18 = 0,75 (mol)

nO (trong O2) = 1,2 + 0,75 = 1,95 (mol)

nO2 = 1,95/2 = 0,975 (mol)

VO2 = 0,975 . 22,4 = 21,84 (l)

Vkk = 21,84 . 5 = 109,2 (l)

nCO2 = 8.8/44 =  0.2 (mol) 

nO2 = 9.6/32 = 0.3 (mol) 

2CO + O2 -to-> 2CO2 

0.2____0.1______0.2

2H2 + O2 -to-> 2H2O 

0.4___0.3-0.1

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

%H2 = 12.5% 

=> D

\(2CO + O_2\xrightarrow{t^o} 2CO_2(1)\\ n_{CO} = n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)\\ n_{O_2(1)} = \dfrac{n_{CO_2}}{2} = 0,1(mol)\\ 2H_2 +O_2 \xrightarrow{t^o} 2H_2O(2)\\ n_{H_2} = 2n_{O_2(2)} = 2.(\dfrac{9,6}{32}-0,1) = 0,4(mol)\\ \Rightarrow \%m_{CO} = \dfrac{0,2.28}{0,2.28 + 0,4.2}.100\% = 87,5\%\\ \%m_{H_2} = 100\% - 87,5\% = 12,5\%\)

PTHH: \(2CO+O_2\underrightarrow{t^o}2CO_2\)  (1)

            \(2H_2+O_2\underrightarrow{t^o}2H_2O\)  (2)

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\\n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{CO}=0,2\left(mol\right)\\n_{H_2}=0,4\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CO}=0,2\cdot28=5,6\left(g\right)\\m_{H_2}=0,4\cdot2=0,8\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{CO}=\dfrac{5,6}{5,6+0,8}\cdot100\%=87,5\%\\\%m_{H_2}=12,5\%\end{matrix}\right.\)