CH4+2O2-to>CO2+2H2O
0,075---0,15
n O2=\(\dfrac{3,36}{22,4}\)=0,15 mol
=>VCH4=0,075.22,4=1,68l
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,075 0,15 ( mol )
\(V_{CH_4}=0,075.22,4=1,68l\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)
CH4 + 2O2 \(\underrightarrow{t^o}\) CO2 + 2H2O
0,075 0,15 ( mol )
\(V_{CH_4}=0,075.22,4=1,68l\)
CH4 (0,075 mol) + 2O2 (0,15 mol) \(\underrightarrow{t^o}\) CO2 + 2H2O.
Thể tích khí metan (đktc) đã tiêu thụ:
V=0,075.22,4=1,68 (lít).