$\text{m$_{C}$=96%.1=0,96(kg) hay 960 g}$

$\text{m$_{S}$=4%.1=0,04(kg) hay 40g}$

$\text{⇒n$_{C}$=960:12=80(mol)}$

$\text{n$_{S}$=40:32=1,25(mol)}$

$\text{PTHH:}$

$\text{C+O$_{2}$$\xrightarrow[]{t^o}$CO$_{2}$}$

$\text{80→80→  80  (mol)}$

$\text{S +   O$_{2}$$\xrightarrow[]{t^o}$SO$_{2}$}$

$\text{1,25→1,25 (mol)}$

$\text{Từ PTHH ta suy ra : n$_{O_2phản ứng}$=80+1,25=81,25(mol)}$

$\text{n$_{CO_2}$=80(mol)}$

$\text{⇒V$_{O_2}$=81,25.22,4=1820(lít)}$

$\text{m$_{CO_2}$=80.44=3520(g) hay 3,52 kg}$

$m_C=1.96\%=0,96(kg);m_S=1-0,96=0,04(kg)$

$\Rightarrow n_C=\dfrac{0,96}{12}=0,08(kmol);n_S=\dfrac{0,04}{32}=0,00125(kmol)$

$PTHH:C+O_2\xrightarrow{t^o}CO_2$

$S+O_2\xrightarrow{t^o}SO_2$

Theo PT: $n_{O_2}=n_C+n_S=0,08125(kmol)=81,25(mol)$

$\Rightarrow V_{O_2(đktc)}=81,25.22,4=1820(lít)$

Theo PT: $n_{CO_2}=n_C=0,08(kmol)$

$\Rightarrow m_{CO_2}=0,08.44=3,52(kg)=3520(g)$