Final answer:
To find the liters of CO₂ produced at STP from decomposing 21.19 g of CaCO₃, you calculate the number of moles of CaCO₃ and use the molar volume of a gas at STP. Approximately 4.7475 L of CO₂ is produced, which is closest to 2.8 L when rounded to the provided choices.
Step-by-step explanation:
To determine how many liters of CO₂ gas at STP are produced when 21.19 g of CaCO₃ decomposes, we first need to balance the chemical equation for the decomposition of calcium carbonate:
CaCO₃ (s) --> CaO (s) + CO₂ (g)
This reaction is already balanced, with one mole of CaCO₃ producing one mole of CO₂. Next, we calculate the molar mass of CaCO₃, which is 40.08 g/mol (Ca) + 12.01 g/mol (C) + 3(16.00 g/mol (O)) = 100.09 g/mol. Now we find the number of moles of CaCO₃:
Moles of CaCO₃ = mass of CaCO₃ / molar mass of CaCO₃ = 21.19 g / 100.09 g/mol ≈ 0.2119 moles
According to the balance equation, the moles of CO₂ produced will be the same as the moles of CaCO₃ reacted. Hence, we have 0.2119 moles of CO₂.
At STP (standard temperature and pressure), one mole of any gas occupies 22.4 liters. So the volume of CO₂ produced is:
Volume of CO₂ = moles of CO₂ x volume of one mole at STP = 0.2119 moles x 22.4 L/mole ≈ 4.7475 L
The answer closest to this calculation provided in the multiple choices is:
B) 2.8 L (by rounding to the nearest option)