Answer to Question #116720 in Chemistry for lin

Question

If 500g of marble are decomposed completely and knowing that it contains 90% calcium carbonate, calculate the mass of calcium oxide and the volume of carbon dioxide ( at S.T.P.) produced.

Answer 1

The molar mass of CaCO₃ is 100 g/mol

The molar mass of CaO is 56 g/mol

The molar volume of a gas at STP is 22.4 L/mol

Solution:

m(CaCO₃) = m(marble) × w = 500 g × 0.9 = 450 g

The balanced chemical equation:

CaCO₃ --> CaO + CO₂

According to the chemical equation: n(CaCO₃) = n(CaO) = n(CO₂)

Moles of CaCO₃ = n(CaCO₃) = Mass of CaCO₃ / Molar mass of CaCO₃

n(CaCO₃) = (450 g) / (100 g/mol) = 4.5 mol

n(CaCO₃) = n(CaO) = n(CO₂) = 4.5 mol

Mass of CaO = Moles of CaO × Molar mass of CaO

m(CaO) = (4.5 mol) × (56 g/mol) = 252 g

m(CaO) = 252 g

Volume of CO₂ = Moles of CO₂ × Molar volume of CO₂

V(CO₂) = (4.5 mol) × (22.4 L/mol) = 100.8 L

V(CO₂) = 100.8 L

Answer:

The mass of calcium oxide (CaO) is 252 g;

The volume of carbon dioxide (CO₂) is 100.8 L.