Question

Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate (CaCO3). Calculate the volume of CO2 at STP produced from the decomposition of 152 g CaCO3 by the reaction.
CaCO3 —> CaO + CO2

Answer 1

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