Em tham khảo nha:

\(\begin{array}{l}
n{O_2} = \dfrac{V}{{22,4}} = \dfrac{{33,6}}{{22,4}} = 1,5\,mol\\
C + {O_2} \xrightarrow{t^0} C{O_2}\\
nC = nC{O_2} = n{O_2} = 1,5\,mol\\
mC = 1,5 \times 12 = 18g\\
VC{O_2} = 1,5 \times 22,4 = 33,6l\\
S + {O_2} \xrightarrow{t^0} S{O_2}\\
nS = nS{O_2} = n{O_2} = 1,5\,mol\\
mS = 1,5 \times 32 = 48g\\
VS{O_2} = 1,5 \times 22,4 = 33,6l\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
nP = \dfrac{{1,5 \times 4}}{5} = 1,2\,mol\\
mP = 1,2 \times 31 = 37,2g\\
n{P_2}{O_5} = \dfrac{{1,2}}{2} = 0,6\,mol\\
m{P_2}{O_5} = 0,6 \times 142 = 85,2g
\end{array}\)

$n_{O_2}=\dfrac{33,6}{22,4}=1,5mol.$

$PTPƯ:C+O_2\xrightarrow{t^o} CO_2$  $(1)$

$Theo$ $pt$ $(1):$ $n_{C}=n_{CO_2}=n_{O_2}=1,5mol.$

$⇒m_{C}=1,5.12=18g.$

$⇒V_{CO_2}=1,5.22,4=33,6l.$

$PTPƯ:S+O_2\xrightarrow{t^o} SO_2$  $(2)$

$Theo$ $pt$ $(2):$ $n_{S}=n_{SO_2}=n_{O_2}=1,5mol.$

$⇒m_{S}=1,5.32=48g.$

$⇒V_{SO_2}=1,5.22,4=33,6l.$

$PTPƯ:4P+5O_2\xrightarrow{t^o} 2P_2O_5$  $(3)$

$Theo$ $pt$ $(3):$ $n_{P}=\dfrac{4}{5}n_{O_2}=1,2mol.$

$⇒m_{P}=1,2.31=37,2g.$

$Theo$ $pt$ $(3):$ $n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=0,6mol.$

$⇒m_{P_2O_5}=0,6.142=85,2g.$

chúc bạn học tốt!