Step 1: Use the proportion method:
44 g of $\text{CO}_2$ (1 mole) takes up 22.4 L at STP. Volume is directly proportional to mass for the same gas.
Step 2: Set up the ratio:
$V = \dfrac{99}{44} \times 22.4$ L. Since $\dfrac{99}{44} = \dfrac{9}{4}$, we get $V = \dfrac{9 \times 22.4}{4}$ L.
Step 3: Calculate:
$22.4 / 4 = 5.6$ L, and $5.6 \times 9 = 50.4$ L.
Step 4: Check the distractors:
A is 1 mole, C is 3 moles and D is 4 moles. Our value of 2.25 moles lies between 2 and 3 moles (44.8 L and 67.2 L), and only 50.4 L fits.
Final Answer:
Using the mole ratio, 99 g of CO2 occupies $50.4$ L at STP.
\[ \boxed{\text{(B) }50.4\ \text{L}} \]