Answer:
The amount of CaCO₃ Jeeves needs to find is approximately 71.392 grams
Step-by-step explanation:
The given parameters are;
The mass of the CaO required = 40 g
The chemical reaction for the production of CaO from CaCO₃ is given as follows;
CaCO₃(s) → CaO(s) + CO₂(g)
Therefore, 1 mole of CaCO₃ produces 1 mole of CaO and 1 mole of CO₂,
The molar mass of CaO = 56.0774 g/mol
Number of moles, n = Mass/(Molar mass)
The number of moles of CaO = 40 g/(56.0774 g/mol) ≈ 0.7133 moles of CaO
Therefore, given that 1 mole of CaO is produced from 1 mole of CaCO₃ 0.7133 moles of CaO will be produced from 0.7133 moles of CaCO₃
The molar mass of CaCO₃ = 100.0869 g/mol
Mass = The number of moles × The molar mass
∴ The mass of CaCO₃ Jeeves needs to find = 0.7133 moles × 100.0869 g/mol ≈ 71.392 g
The mass of CaCO₃ Jeeves needs to find ≈ 71.392 grams.