Answer:
V = 5.74 L
Explanation:
Given data:
Volume of carbon dioxide produced = ?
Temperature and pressure = standard
Mass of calcium carbonate = 25.6 g
Solution:
Chemical equation:
CaCO₃ → CaO + CO₂
Number of moles of calcium carbonate:
Number of moles = mass / molar mass
Molar mass of calcium carbonate = 100 g/mol
Number of moles = 25.6 g/ 100 g/mol
Number of moles = 0.256 mol
Now we will compare the mole of calcium carbonate with carbon dioxide.
CaCO₃ : CO₂
1 : 1
0.256 : 0.256
Volume of CO₂:
PV = nRT
R = general gas constant = 0.0821 atm/L / mol.K
By putting values,
1 atm× V = 0.256 mol × 0.0821 atm/L / mol.K× 273.15 K
V = 5.74 atm/L / 1 atm
V = 5.74 L
