2) PTHH: C+O2=to=> CO2(1)
4P+5O2=to=>2P2O5(2)
Từ (1), (2), ta có:
\(n_{O_2\left(líthuyet\right)}=n_C+\frac{5}{4}.n_P=\frac{2,4}{12}+\frac{5}{4}.\left(\frac{9,3}{31}\right)=0,575mol\)
\(n_{O_2\left(thucte\right)}=\frac{100}{90}.n_{O_2\left(líthuyet\right)}=\frac{100}{90}.0,575=\frac{23}{36}mol\)
\(V_{O_2\left(thucte\right)}=\frac{23}{36}.22,4\simeq14,311l\)
b) \(m_{CO_2}=0,2.44=8,8g\)
\(n_{P_2O_5}=0,3.\left(\frac{2}{4}\right)=0,15mol\Rightarrow m_{P_2O_5}=0,15.142=21,3g\)
\(\%CO_2=\frac{8,8}{8,8+21,3}=29,24\%\)
\(\%P_2O_5=100\%-29,24\%=70,76\%\)