Bài 11:
a, Gọi: \(\left\{{}\begin{matrix}n_{C_2H_2}=x\left(mol\right)\\n_{C_3H_8}=y\left(mol\right)\end{matrix}\right.\)
Có: dhh/H2 = 15,25
\(\Rightarrow\dfrac{26x+44y}{x+y}=15,25.2\) \(\Rightarrow x=3y\)
%V cũng là % số mol.
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{x}{x+y}.100\%=\dfrac{3y}{3y+y}.100\%=75\%\\\%V_{C_3H_8}=25\%\end{matrix}\right.\)
b, Ta có: \(n_{hh}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_2}=0,2.75\%=0,15\left(mol\right)\\n_{C_3H_8}=0,05\left(mol\right)\end{matrix}\right.\)
PT: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
_____0,15__0,375 (mol)
\(C_3H_8+5O_2\underrightarrow{t^o}3CO_2+4H_2O\)
0,05___0,25 (mol)
\(\Rightarrow V_{O_2}=22,4.\left(0,375+0,25\right)=14\left(l\right)\)
