Answer:
V = 1.57 L
Explanation:
Given data:
Mass of iron(III) carbonate = 50.0 g
Volume of CO₂ formed = ?
Temperature = standard = 273 K
Pressure = standard = 1 atm
Solution:
Chemical equation:
Fe₂(CO₃)₃ → Fe₂O₃ + 3CO₂
Number of moles of Fe₂(CO₃)₃ :
Number of moles = mass/molar mass
Number of moles = 50.0 g /235.8 g/mol
Number of moles = 0.21 mol
Now we will compare the moles of Fe₂(CO₃)₃ and CO₂.
Fe₂(CO₃)₃ : CO₂
1 : 3
0.21 : 1/3×0.21 = 0.07 mol
Volume of CO₂:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
By putting values,
1 atm × V =0.07 mol × 0.0821 atm.L/mol.K × 273 K
V = 1.57 atm.L / 1 atm
V = 1.57 L
