gọi a là số mol C2H2
b là số mol C3H8
Ta có
\(a+b\)=\(\frac{4,48}{22,4}\)=0,2(mol)(1)
\(\text{26a+44b=15,25.2(a+b)}\)
\(\rightarrow\)13,5b-4,5a=0(2)
(1)(2)\(\Rightarrow\)a=0,15 b=0,05
2C2H2+5O2\(\rightarrow\)4CO2+2H2O(3)
C3H8+5O2\(\rightarrow\)3CO2+4H2O(4)
\(\rightarrow\)nO2=nO2(3)+nO2(4)=0,15.\(\frac{5}{2}\)+0,05.5=0,625(mol)
\(\rightarrow\)VO2=\(\text{0,625.22,4=14(l)}\)