Molecular equation: Ca(OH)2 + CO2 = CaCO3 + H2O

Balanced equation: Ca(OH)2 + CO2 = CaCO3 + H2O

Now we need to calculate the amount of CaCO3 formed when 12 g of calcium hydroxide reacts with 27 L of carbon dioxide.

Calculate the moles of Ca(OH)2:
Molar mass of Ca(OH)2 = 40.08 g/mol + 2(1.008 g/mol) + 2(16.00 g/mol) = 74.08 g/mol
Moles of Ca(OH)2 = 12 g / 74.08 g/mol = 0.162 mol

Calculate the limiting reactant (the reactant that will be completely used up):
First, convert 27 L of CO2 to moles using the ideal gas law:
P(V) = nRT
n = PV/RT = (1 atm)(27 L) / (0.0821 L.atm/mol.K)(298 K) = 0.990 mol

Determine the ratio of reactants (1:1 from the balanced equation):
Ca(OH)2 : CO2 = 1 : 1

Since the moles of Ca(OH)2 (0.162 mol) is less than the moles of CO2 (0.990 mol), Ca(OH)2 is the limiting reactant.

Calculate the moles of CaCO3 formed:
Moles of CaCO3 = 0.162 mol

Calculate the mass of CaCO3 formed:
Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + 3(16.00 g/mol) = 100.09 g/mol
Mass of CaCO3 formed = 0.162 mol * 100.09 g/mol = 16.21 g

Therefore, 16.21 g of calcium carbonate is formed in the reaction. Calcium hydroxide will be in excess.