\(n_{C_2H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{^{t^0}}2CO_2+H_2O\)
\(0.1........0.25...........0.2\)
\(V_{O_2}=0.25\cdot22.4=5.6\left(l\right)\)
\(V_{CO_2}=0.2\cdot22.4=4.48\left(l\right)\)
n C2H2 = 22,4/22,4 = 1(mol)
$2C_2H_2 + 5O_2 \xrightarrow{t^o} 4CO_2 + 2H_2O$
Theo PTHH :
n O2 = 5/2 n C2H2 = 2,5(mol) => V O2 = 2,5.22,4 = 56(lít)
n CO2 = 2n C2H2 = 2(mol) => V CO2 = 2.22,4 = 44,8(lít)