Answer:
110 g
Explanation:
Step 1: Write the balanced equation for the decomposition of calcium carbonate
CaCO₃(s) ⇒ CaO(s) + CO₂(g)
Step 2: Convert 260.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15 = 260.0 + 273.15 = 533.2 K
Step 3: Calculate the moles of CO₂ produced
We will use the ideal gas equation.
P × V = n × R × T
n = P × V/R × T = 1 atm × 48.0 L/0.0821 atm.L/mol.K × 533.2 K = 1.10 mol
Step 4: Calculate the moles of CaCO₃ that reacted to produce 1.10 moles of CO₂
The molar ratio of CaCO₃ to CO₂ is 1:1. The moles of CaCO₃ that reacted are 1/1 × 1.10 mol = 1.10 mol.
Step 5: Calculate the mass corresponding to 1.10 moles of CaCO₃
The molar mass of CaCO₃ is 100.09 g/mol.
1.10 mol × 100.09 g/mol = 110 g
