Final answer:
After 25.0 mL of dry ice sublimes, it would produce approximately 19.84 L of CO₂ gas at Standard Temperature and Pressure (STP).
Step-by-step explanation:
To determine how many liters (L) of CO₂(g) are present after 25.0 mL of dry ice sublimes, we employ the principles of the ideal gas law and stoichiometry. First, we need to calculate the number of moles of CO₂ that 25.0 mL of dry ice, or solid CO₂, would produce upon sublimation.
Assuming the density of dry ice is approximately 1.56 g/mL, the mass of 25.0 mL of dry ice is 25.0 mL × 1.56 g/mL = 39.0 g. Now, using the molecular weight of CO₂, which is about 44.01 g/mol, the number of moles of CO₂ is 39.0 g / 44.01 g/mol, which equals approximately 0.886 moles of CO₂.
At Standard Temperature and Pressure (STP), which is 0°C and 1 atmosphere, 1 mole of an ideal gas occupies 22.4 L. Therefore, if we assume STP conditions, the volume of CO₂ that sublimes from 25.0 mL of dry ice can be calculated as 0.886 moles × 22.4 L/mol, which is approximately 19.84 L of CO₂ gas.