C2H5Oh+3O2-to>2CO2+3H2O
0,05-------0,15------0,1
nC2H5OH=2,3\46=0,05 mol
=>VO2=22,4.0,15=3,36l
=>VCO2=22,4.0,1=2,24l
a) \(C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)
b) \(n_{C2H5OH}=\frac{23}{46}=0,05mol\)
Theo PTHH : \(n_{O2}=3n_{C2H5OH}=3.0,05=0,15mol\)
=> \(V_{O2}=22,4.0,15=3,36l\)
Theo PTHH (tương tự bn tự tính nha) \(n_{CO2}=0,1mol\)
=> \(V_{CO2}=22,4.0,1=2,24l\)