Question

Write a balanced chemical equation, including physical state symbols, for the decomposition of solid calcium carbonate (CaCO₃) into solid calcium oxide and gaseous carbon dioxide. Suppose 23.0 L of carbon dioxide gas are produced by this reaction, at a temperature of 380.0 °C and pressure of exactly 1 atm. Calculate the mass of calcium carbonate that must have reacted. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

CaCO3 (s)  → CaO (s) + CO2 (g)

The mass of carbonate that must have reacted was 43.03 grams

Explanation:

CaCO3  → CaO + CO2

Relation between reactant and product is 1:1

Let's apply the Ideal Gas Law to find out the moles of CO2 which were produced.

P . V = n . R . T

1 atm . 23 L = n . 0.082 L.atm/mol.K . 653K

(1atm . 23L) / (0.082 mol.K/L.atm . 653K) = n

0.43 moles = n

0.43 moles of CO2, were produced from 0.43 moles of CaCO3.

Molar weight of CaCO3 = 100.08 g/m

Mass = Molar weight . moles

Mass = 100.08 g/m 0.43 m = 43.03 g