A 14.584 g sample of CaCl2 was added to 12.125 g of K2CO3 and mixed in water. What is the limiting reactant and what is the theoretical yield of CaCO3

The correct answer is 8.786 g CaCO₃

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.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24