nC2H2= 0,2(mol)
PTHH: 2 C2H2 + 5 O2 -to-> 4 CO2 + 2 H2O
nO2= 5/2 x 0,2=0,5(mol)
=> V(O2,ddktc)=0,5.22,4=11,2(l)
pt: \(2C_2H_2+5O_2\rightarrow4CO_2+2H_2O\)
Theo pt: \(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=\dfrac{5}{2}.\dfrac{5,2}{26}=0,5mol\)
\(\Rightarrow V_{O_2}=0,5.22.4=11,2lit\)
\(n_{C_2H_2} = \dfrac{5,2}{26} = 0,2(mol)\\ C_2H_2 + \dfrac{5}{2}O_2 \xrightarrow{t^o} 2CO_2 + H_2O\\ n_{O_2} = \dfrac{5}{2}n_{C_2H_2} = 0,5(mol\\ \Rightarrow V_{O_2} = 0,5.22,4 = 11,2(lít)\)