Be careful, its kinetics!

The balance reaction should be $\ce{2SO2 + O2 -> 2SO3}$ . This means that $1$ mol of $\ce{O2}$ gives $2$ mol of $\ce{SO3}$ . Since $8 \text{ g/s } = \frac{1}{4} \text{ mol/s}$ of $\ce{O2}$ disappearance, it gives $\frac{1}{2} \text{ mol/s} = \boxed{40 \text{ g/s}}$ of $\ce{SO3}$ appearance.

$2S{ O }_{ 2 }\quad +{ \quad O }_{ 2 }\quad \rightarrow \quad 2S{ O }_{ 3 }\\ According\quad to\quad Kinetics\\ \frac { -1 }{ 2 } \frac { d[S{ O }_{ 2 }] }{ dt } =-\frac { d[{ O }_{ 2 }] }{ dt } =\frac { 1 }{ 2 } \frac { d[S{ O }_{ 3 }] }{ dt } \quad ---------(1)\\ Given\quad -\frac { d[{ O }_{ 2 }] }{ dt } =8gm/s=\frac { 8 }{ 32 } moles/s=\frac { 1 }{ 4 } moles/s\\ Now\quad Using\quad Equation\quad ---------(1)\\ \frac { 1 }{ 2 } \frac { d[S{ O }_{ 3 }] }{ dt } =\frac { 1 }{ 4 } moles/s\\ \frac { d[S{ O }_{ 3 }] }{ dt } =\frac { 1 }{ 2 } moles/s=\frac { 32+16+16+16 }{ 2 } gm/s=40gm/s\\ \\$

@Akhil Bansal kindly post ur solution, I accidentally answered it correctly.