Answer:
V = 34.1 L
Explanation:
Given data:
Mass of CaCO₃ = 152 g
Volume of CO₂ produced = ?
Solution:
Chemical equation:
CaCO₃ → CaO + CO₂
Number of moles of CaCO₃ :
Number of moles = mass/ molar mass
Number of moles = 152 g / 100 g/mol
Number of moles = 1.52 mol
Now we will compare the moles of CaCO₃ with CO₂
CaCO₃ : CO₂
1 : 1
1.52 : 1.52
Volume of CO₂:
PV = nRT
Standard temperature = 273 K
Standard pressure = 1 atm
R = 0.0821 atm.L/mol.K
Now we will put the value sin formula:
V = nRT/P
V = 1.52 mol × 0.0821 atm.L/mol.K × 273 K / 1 atm
V = 34.1 L
