Final answer:
To calculate the mass of CaO produced from 1.0 kg of limestone, use stoichiometry and the molar mass of CaCO3. The mass of CaO produced is 560 g. The percentage yield is 89.3%.
Step-by-step explanation:
To solve this problem, we need to use stoichiometry and the given information from the balanced equation:
CaCO3(s) → CaO(s) + CO2(g)
The molar mass of CaCO3 is 100.09 g/mol. We can calculate the moles of CaCO3 in 1.0 kg by dividing the mass by the molar mass:
1.0 kg = 1000 g
Moles of CaCO3 = (1000 g / 100.09 g/mol) = 9.997 mol ≈ 10.0 mol
From the balanced equation, we know that 1 mole of CaCO3 produces 1 mole of CaO. So, the moles of CaO produced will be the same as the moles of CaCO3:
Moles of CaO = 10.0 mol
To calculate the mass of CaO produced, we need to multiply the moles by the molar mass of CaO:
Mass of CaO = (10.0 mol) x (56.08 g/mol) = 560 g
The percentage yield is calculated by dividing the actual yield (0.5 kg or 500 g in this case) by the theoretical yield (560 g) and multiplying by 100:
Percentage yield = (500 g / 560 g) x 100 = 89.3%