Given the balanced equation that follows, how many milliliters of carbon dioxide gas at STP are
produced from the decomposition of 3.67 g of ferric carbonate?
Fe2(CO3)3 - 'Fe2O3 + 3C02
1,000 mL = 1 liter
Molar mass of ferric carbonate (Fe2(CO3)3) = 291.7 g/mol

870 mL

Explanation:

Given data:

Mass of ferric carbonate  = 3.67 g

Molar mass of ferric carbonate  = 291.7 g/mol

Volume of CO₂ 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

it's filtration, but this is an assumption so my apologies if i got this

it is the most commonly accepted one