Answer:
The correct answer is 8.786 g CaCO₃
Step-by-step explanation:
The balanced reaction is the following:
CaCl₂(ac) + K₂CO₃(ac) → CaCO₃(s) + 2 KCl(ac)
From the stoichiometry, 1 mol of CaCl₂ (111 g) reacts with 1 mol of K₂CO₃ (138 g) to form 1 mol CaCO₃(100 g) and 2 moles of KCl (149 g).
The stoichiometric ratio CaCl₂/K₂CO₃ is: 111 g/138 g= 0.80 g CaCl₂/K₂CO₃.
We have 14.584 g CaCl₂ and 12.125 g K₂CO₃, which gives a ratio of: 14.584g/12.125 g= 1.2 g CaCl₂/K₂CO₃.
0.8 ∠ 1.2 ⇒ K₂CO₃ is the limiting reactant
We use the limiting reactant to calculate the grams of CaCO₃ produced. For this, we know that from 138 g K₂CO₃ 100 g of CaCO₃ are produced. So, we multiply the amount of K₂CO₃ by this stoichiometric ratio to obtain the grams of CaCO₃ produced:
12.125 g K₂CO₃ x 100 g CaCO₃/138 g K₂CO₃= 8.786 g CaCO₃
Therefore, the theoretical yield of CaCO₃ is 8.786 g.