Phản ứng:
\(C_2H_5+3O_2\rightarrow2CO_2+3H_2O\)
Ta có :
\(V_{C2H5OH}=25.45\%=11,25\left(ml\right)\)
\(\Rightarrow m_{C2H5OH}=11,25.0,8=9\left(g\right)\)
\(\Rightarrow n_{C2H5OH}=\frac{9}{46}\left(mol\right)\)
\(n_{O2}=3n_{C2H5OH}=\frac{9}{46}.3=\frac{27}{46}\left(mol\right)\)
\(\Rightarrow V_{O2}=\frac{27}{46}.22,4=13,148\left(l\right)\)
\(\Rightarrow V_{kk}=13,148.5=65,74\left(l\right)\)
\(n_{CO2}=2n_{C2H5OH}=\frac{9}{46}.2=\frac{9}{23}\left(mol\right)\)
\(\Rightarrow m_{CO2}=\frac{9}{23}.44=17,217\left(g\right)\)
a)\(C2H6O+3O2-->2CO2+3H2O\)
b)\(m_{C2H6O}=25.0,8=20\left(g\right)\)
\(n_{C2H6O}=\frac{20}{46}=\frac{5}{23}\left(mol\right)\)
\(nO2=3n_{C2H6O}=\frac{15}{23}\left(mol\right)\)
\(V_{O2}=\frac{15}{23}.22,4=14,6\left(l\right)\)
\(V_{kk}=5V_{O2}=14,6.5=73\left(l\right)\)
c)\(n_{CO2}=2n_{C2H6O}=\frac{10}{23}\left(mol\right)\)
\(m_{CO2}=\frac{10}{23}.44=19,13\left(g\right)\)
