\(n_{CH_4}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(PTHH:CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
(mol)____0,25___0,5____0,25____0,5__
\(V_{O_2}=0,5.22,4=11,2\left(l\right)\)
\(V_{KK}=5.V_{O_2}=5.11,2=56\left(l\right)\)
a)nCH4=\(\frac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH:
CH4+ 2O2 \(\underrightarrow{t^o}\)CO2+2H2O
0,25→0,5→0,25→0,5(mol)
b) VO2= 0,5.22,4= 11,2(l)
c) Vkk= 5.11,2=56(l)