Answer
$m_{O_2}=0.998 kg$ and $m_{SO_2}=1.998 kg$
Work Step by Step
The chemical reaction is given as:
$S+O_2 \rightarrow SO_2$
The molecular weight of $So_2$ is: $M_{So_2}=M_S+M_{o_2}=32.065+32=64.065$kg/kmol
Using molar mass, we balance equation per unit kmol of the chemical reaction as:
$\dfrac{m_{O_2}}{m_{S}}=\dfrac{M_{O_2}}{M_{S}}=\dfrac{32}{32.065}$ (here $m_S= 1 kg$)
This gives
$m_{O_2}=0.998 kg$
Now, we have
$\dfrac{m_{SO_2}}{m_{S}}=\dfrac{M_{O_2}}{M_{S}}=\dfrac{32}{32.065}$ (here $m_S= 1 kg$)
This gives
$m_{SO_2}=1.998 kg$
