What is the volume of the produced SO2 gas at STP?
You have an excess of oxygen, right...?

So the limiting reactant is the sulfur, which is $2.4$% of the coal. 

Spoiler:
Step 1:


   So calculate the sulfur content, which is $3.1 \times 10^7$ kilograms multiplied by $2.4 \times 10^{-2}$ which is $7.44 \times 10^5$ kilograms of sulfur. 


Step 2:


   The $\ce{SO_2}$ is $50$% sulfur by mass (32 amu of sulfur vs. 64 amu total) so there is $1.488 \times 10^6$ kilograms of $\ce{SO_2}$ produced. 


Step 3 (for moles, optional):


   If you want moles, divide by $64$, which is $2.325 \times 10^4$ moles (someone check my math)


Then again, you emphasized STP, so proceed with caution (sulfur doesn't react at $20$ degrees Celsius).