a) Gọi số mol C2H2, C4H10 là a, b (mol)
=> \(\left\{{}\begin{matrix}a+b=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\\overline{M}=\dfrac{26a+58b}{a+b}=25,8.2=51,6\left(g/mol\right)\end{matrix}\right.\)
=> a = 0,01 (mol); b = 0,04 (mol)
b)
\(\left\{{}\begin{matrix}n_{O_2}=\dfrac{33,6.20\%}{22,4}=0,3\left(mol\right)\\n_{N_2}=\dfrac{33,6.80\%}{22,4}=1,2\left(mol\right)\end{matrix}\right.\)
- Giả sử X cháy hết
PTHH: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
0,01-->0,025-->0,02
\(2C_4H_{10}+13O_2\underrightarrow{t^o}8CO_2+10H_2O\)
0,04--->0,26--->0,16
=> \(n_{O_2\left(pư\right)}=0,025+0,26=0,285\left(mol\right)< 0,3\left(mol\right)\)
=> Điều giả sử là đúng
Y chứa \(\left\{{}\begin{matrix}O_{2\left(dư\right)}:0,3-0,285=0,015\left(mol\right)\\N_2:1,2\left(mol\right)\\CO_2:0,02+0,16=0,18\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%V_{O_2}=\dfrac{0,015}{0,015+1,2+0,18}.100\%=1,075\%\\\%V_{N_2}=\dfrac{1,2}{0,015+1,2+0,18}.100\%=86,022\%\\\%V_{CO_2}=\dfrac{0,18}{0,015+1,2+0,18}.100\%=12,903\%\end{matrix}\right.\)
