Final answer:
To produce 10 grams of CaCO₃, a theoretical amount of approximately 11.098 grams of CaCl₂ is required, using molar mass and stoichiometry to calculate the correct ratio based on the 1:1 mole relationship in the reaction.
Step-by-step explanation:
To determine the maximum (theoretical) amount of CaCO₃ that can be produced from a precipitation reaction, consider the amount of reactants you're starting with and use stoichiometry to relate it to the desired product.
Given the balanced chemical reaction CaCl₂ + H₂CO₃ → CaCO₃ + 2HCl, if you want to produce 10 g of calcium carbonate, first, you need to know the molar mass of CaCl₂ and CaCO₃. The molar mass of CaCl₂ is approximately 110.98 g/mol, while that of CaCO₃ is approximately 100.09 g/mol.
The reaction shows a 1:1 mole ratio between CaCl₂ and CaCO₃, which means 1 mole of calcium chloride will produce 1 mole of calcium carbonate.
Applying the molar ratio and stoichiometry, 10 g of CaCO₃ (which is ≈ 0.1 moles, since 10 g ÷ 100.09 g/mol ≈ 0.1) would need approximately 0.1 moles of CaCl₂.
Therefore, multiply the molar amount of CaCO₃ (0.1) by the molar mass of CaCl₂ (110.98 g/mol) to obtain the required mass of calcium chloride, resulting in ≈ 11.098 g of CaCl₂ needed to theoretically produce 10 g of CaCO₃.