Question

[Mr. Phong]

giúp mình câu 8 với

Answer 1

[Lê Ng Hải Anh]

Câu 8:

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

Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=\dfrac{20,8}{32}=0,65\left(mol\right)\left(2\right)\)

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

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,2.22,4}{6,72}.100\%\approx66,67\%\\\%V_{C_2H_2}\approx33,33\%\end{matrix}\right.\)

b, Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=0,4\left(mol\right)\Rightarrow V_{CO_2}=0,4.22,4=8,96\left(l\right)\)