Gọi \(n_{H2}=x\left(mol\right)\), \(n_{C2H2}=y\left(mol\right)\)
Ta có: \(n_X=x+y=\dfrac{17,92}{22,4}=0,8\left(mol\right)\)(1)
\(d_X\)/N2 =0,5\(\Rightarrow M_X=0,5\cdot28=14\left(g\right)\)
Mà \(M_X=\dfrac{m_X}{n_X}=\dfrac{2x+26y}{0,5}=14\)
\(\Rightarrow2x+26y=14\cdot0,8=11,2\left(g\right)\)(2)
Từ(1),(2)\(\Rightarrow\left\{{}\begin{matrix}2x+26y=11,2\\x+y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,4\left(mol\right)\\y=0,4\left(mol\right)\end{matrix}\right.\)
PTHH:\(2C_2H_2+O_2\rightarrow4CO_2+2H_2O\)
0,4 1 (mol)
\(2H_2+O_2\rightarrow2H_2O\)
0,4 0,2 (mol)
Theo PTHH(1),(2): \(n_{O2}\)p/ứ = 1+0,2=1,2(mol)
Theo bài ra: nO2 p/ứ =\(\dfrac{35,84}{22,4}=1,6\left(mol\right)\)
=> nO2 dư =1,6-1,2=0,4(mol)
=> Trong Y gồm CO2 và O2 dư
Theo PTHH(1):\(n_{CO2}=2.n_{C2H2}2\cdot0,4=0,8\left(mol\right)\)
Mà % về thể tích cũng là % về số mol nên:
\(\Rightarrow\%V_{O2}=\dfrac{0,4}{0,4+0,8}\cdot100\%\approx33,33\%\)
\(\%V_{CO2}=100\%-33,33\%=66,67\%\)
\(\Rightarrow\%m_{O2}=\dfrac{0,4\cdot32}{0,4\cdot32+0,8\cdot44}\cdot100\%\approx26,7\%\)
\(\%m_{CO2}=100\%-26,7\%=73,3\%\)