Ta có PTHH : CH4+2O2--->CO2+2H2O
theo PT: 1 2 1 1 (mol)
theo bài: 0,2--->0,4------>0,2 (mol)
nCH4=4,48/22,4=0,2(mol);
VO2=0,4.22,4=8,96(l)
VCO2=0,2.22,4=4,48 (l)
\(n_{CH_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Theo PT ta có: \(n_{O_2}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,4.22,4=8,96\left(l\right)\)
Theo PT ta có: \(n_{CO_2}=n_{CH_4}=0,2\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
\(n_{CH_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: \(CH_4+2O_2\underrightarrow{t^0}CO_2+2H_2O\)
Theo PTHH: \(n_{CH_4}=n_{O_2}=1:2\)
\(\Rightarrow n_{O_2}=n_{CH_4}.2=0,2.2=0,4\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,4.22,4=8,96\left(l\right)\)
Theo PTHH: \(n_{CH_4}=n_{CO_2}=1:1\Rightarrow n_{CH_4}=n_{CO_2}=0,2\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
Bạn nên ghi rõ là ĐKTC hay ĐKT nhé, dưới đây tớ giải cả hai vì tớ ko bik cái nào.
PTHH: CH4 + 2O2 -> CO2 + 2H2O
TH1: Khi CH4 ở ĐKTC:
\(n_{CH_4}\left(ĐKTC\right)=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo pt: \(n_{O_2}=n_{CH_4}.\dfrac{2}{1}=0,4\left(mol\right)\)
\(V_{O_2}\left(ĐKTC\right)=0,4.22,4=8.96\left(l\right)\)
\(V_{O_2}\left(ĐKT\right)=0,4.24=9,6\left(l\right)\)
\(n_{CO_2}=0,2\left(mol\right)\)
\(V_{CO_2}\left(ĐKTC\right)=0,2.22,4=4,48\left(l\right)\)
\(V_{CO_2}\left(ĐKT\right)=0,2.24=4,8\left(l\right)\)
TH2: Khi CH4 ở ĐKT: PTHH như trên đầu ko thay đổi
\(n_{CH_4}\left(ĐKT\right)=\dfrac{4,48}{24}=0,18\left(mol\right)\)
Theo pt: \(n_{O_2}=n_{CH_4}.\dfrac{2}{1}=0,18.\dfrac{2}{1}=0.36\left(mol\right)\)
\(V_{O_2}\left(ĐKTC\right)=0,36.22,4=8,064\left(l\right)\)
\(V_{O_2}\left(ĐKT\right)=0,36.24=8,64\left(l\right)\)
\(n_{CO_2}=n_{CH_4}=0,18\left(mol\right)\)
\(V_{CO_2}\left(ĐKTC\right)=0,18.22,4=4,032\left(l\right)\)
\(V_{CO_2}\left(ĐKT\right)=0,18.24=4,32\left(l\right)\)
