Answer:
870 mL
Explanation:
Given data:
Mass of ferric carbonate  = 3.67 g
Molar mass of ferric carbonate  = 291.7 g/mol
Volume of COâ‚‚ produced = ?
Solution:
Chemical equation:
Fe₂(CO₃)₃  →  Fe₂O₃ + 3CO₂
Number of moles of ferric carbonate:
Number of moles = mass / molar mass
Number of moles = 3.67 g/291.7 g/mol
Number of moles = 0.013 mol
Now we will compare the moles of ferric carbonate with carbon dioxide.
                Fe₂(CO₃)₃     :      CO₂
                  1           :       3
                  0.013       :      3×0.013 = 0.039 mol
Litters of carbon dioxide at STP:
PV = nRT
V = nRT/P
V = 0.039 mol × 0.0821 atm.L/mol.K × 273 K/ 1 atm
V = 0.87 L
Litter to mL:
1000 mL = 1 L
0.87L×1000 mL/1 L = 870 mL
