Answer:
72.8 liters (see comment below)
Step-by-step explanation:
You want the volume of CO₂ at STP resulting from the decomposition of 3.25 moles of CaCO₃.
Reaction
The balanced reaction is ...
CaCO₃(s) + heat ⇒ CaO(s) + CO₂(g)
This tells you that 3.25 moles of CaCO₃ will evolve 3.25 moles of CO₂ gas.
Volume
At STP, each mole of gas has a volume of 22.4 liters. The volume of CO₂ produced is ...
(3.25 mol) × (22.4 L/mol) ≈ 72.8 L
At STP, the volume of gas produced is 72.8 liters.
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Additional comment
After 1982, the IUPAC defined STP as 0 °C, 100 kPa, so one mole of gas occupies 22.711 liters. If this is the value of STP you're to use, then the appropriate answer is ...
(3.25 mol) × (22.711 L/mol) ≈ 73.8 L
Prior to 1982, IUPAC standard pressure was 1 atm, or 101.325 kPa. At that pressure the volume of one mole is 22.414 liters. Many authors continue to use this definition of STP, so that is the value we have used in the answer above.
We have rounded the volume to 3 significant figures, in accordance with the precision of the given amount of material.