Theo de bai ta co
Khoi luong moi chat trong1 mol hon hop la
mC4H10=\(\dfrac{80\%.20}{100\%}=16g\)
mCH4=20-16=4g
\(\Rightarrow\)nC4H10=\(\dfrac{16}{58}\approx0,3mol\)
nCH4=\(\dfrac{4}{16}=0,25mol\)
Ta co pthh
2C4H10 + 13O2 \(\rightarrow\)8CO2 + 10H2O
CH4 + 2O2 \(\rightarrow\)CO2 + 2H2O
Theo Pthh 1
nCO2=\(\dfrac{8}{2}nC4H10=\dfrac{8}{2}.0,3=1,2mol\)
Theo pthh 2
nCO2=nCH4=0,25 mol
\(\Rightarrow\)Khoi luong cua CO2 sau phan ung la
mCO2=(1,2+0,25).44=63,8g
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