Question

a) when 2 moles of potassium carbonate and 1 mole of calcium chloride are combined is a limiting reagent and is a reagent in excess.
b) when 0.0375 mole of potassium carbonate and 0.122 mole of calcium chloride are combined is a limiting reagent and is a reagent in excess.

Explanation:

Step1: Write the balanced chemical equation

The reaction between potassium carbonate ($K_2CO_3$) and calcium chloride ($CaCl_2$) is $K_2CO_3 + CaCl_2=CaCO_3\downarrow + 2KCl$. The mole - ratio of $K_2CO_3$ to $CaCl_2$ is 1:1.

Step2: Determine limiting and excess reagents for part a

Given 2 moles of $K_2CO_3$ and 1 mole of $CaCl_2$. According to the 1:1 mole - ratio, for every 1 mole of $CaCl_2$, 1 mole of $K_2CO_3$ reacts. Since we have 1 mole of $CaCl_2$, it will react with 1 mole of $K_2CO_3$. So, $CaCl_2$ is the limiting reagent and $K_2CO_3$ is the reagent in excess.

Step3: Determine limiting and excess reagents for part b

Given 0.0375 mole of $K_2CO_3$ and 0.122 mole of $CaCl_2$. According to the 1:1 mole - ratio, 0.0375 mole of $K_2CO_3$ will react with 0.0375 mole of $CaCl_2$. Since 0.0375 < 0.122, $K_2CO_3$ is the limiting reagent and $CaCl_2$ is the reagent in excess.

Answer:

a) $CaCl_2$, $K_2CO_3$
b) $K_2CO_3$, $CaCl_2$