The question has missing information. At part 1 it is "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."
Part 2. "Suppose 19.0 L of carbon dioxide gas are produced by this reaction, at a temperature of 290.0°C and pressure of exactly 1 atm. Calculate the mass of calcium carbonate that must have reacted (...)"
Answer:
41.0 g
Explanation:
1. Calcium oxide has molecular formula CaO and carbon dioxide COâ‚‚, thus, the reaction will be:
CaCO₃(s) → CaO(s) + CO₂(g)
The equation is already balanced because there's the same number of each element on both sides.
2. First, let's calculate the number of moles of COâ‚‚ produced by the ideal gas law:
PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm.L/mol.K), and T is the temperature (290°C = 273 = 563 K).
1*19 = n*0.082*563
46.166n = 19
n = 0.4116 mol
By the stoichiometry of the reaction:
1 mol of CaCO₃ ------ 1 mol of CO₂
x ----- 0.4116 mol
By a simple direct three rule:
x = 0.4116 mol of CaCO₃.
The molar mass of the calcium carbonate is 100 g/mol, thus the mass (m) is the number of moles multiplied by it:
m = 0.4116*100
m = 41.16 g = 41.0 g
