Question

Calcium carbonate (10 g) reacts with aqueous HCl (0.3 mol) according to the reaction, $CaC{{O}_{3}}(s)+2HCl(aq)\to CaC{{l}_{2}}(aq)+C{{O}_{2}}(g)+{{H}_{2}}O(l)$. The number of moles of $CaC{{l}_{2}}$ formed in the reaction is:
(A) 0.1 mol
(B) 0.2 mol
(C) 0.01 mol
(D) 0.066 mol

Answer 1

$CaC{{O}_{3}}$ is the limiting reagent in the given reaction. First calculate the number of moles of $CaC{{O}_{3}}$ that are reacting with HCl. The actual amount of $CaC{{O}_{3}}$ reacted with HCl will determine the amount of the $CaC{{l}_{2}}$.

Complete step by step answer:
The given chemical reaction is
$CaC{{O}_{3}}(s)+2HCl(aq)\to CaC{{l}_{2}}(aq)+C{{O}_{2}}(g)+{{H}_{2}}O(l)$

Given mass of calcium carbonate ($CaC{{O}_{3}}$) that reacts with 0.3 mole of HCl = 10 g.
The number of moles $CaC{{O}_{3}}$ will be calculated as,
$n=\dfrac{\text{given mass of CaC}{{\text{O}}_{3}}}{\text{molar mass of CaC}{{\text{O}}_{3}}}$

Let us first calculate the molar mass of $CaC{{O}_{3}}$.
Molar mass of $CaC{{O}_{3}}$ = $40+12+16\times 3=100g\,mo{{l}^{-1}}$
Then, the number of moles present in 10 g of $CaC{{O}_{3}}$ will be equal to
$n=\dfrac{10g}{100g\,mo{{l}^{-1}}}=0.1$ mol

From the above reaction, we can see that one mole of $CaC{{O}_{3}}$ reacts with two moles of HCl.
1 mole of $CaC{{O}_{3}}$ = 2 moles of HCl

Then, dividing both sides by 10, we have
0.1 mole of$CaC{{O}_{3}}$ = 0.2 mole of HCl
0.3 mole of HCl is present in the reaction and we can clearly see that for only 0.2 mole of HCl, $CaC{{O}_{3}}$ is consumed completely. Hence, $CaC{{O}_{3}}$ is a limiting reagent as it limits the yield of calcium chloride ($CaC{{l}_{2}}$). The reaction cannot proceed further to form $CaC{{l}_{2}}$ without $CaC{{O}_{3}}$.

From the above chemical reaction, we see that two moles of HCl forms one mole of calcium chloride $CaC{{l}_{2}}$, i.e.
2 moles of HCl = 1 mole of $CaC{{l}_{2}}$
Or we can also write as, 1 mole of HCl = $\dfrac{1}{2}$ mole of $CaC{{l}_{2}}$
But since only 0.2 mole of HCl has reacted with $CaC{{O}_{3}}$.
Thus, 0.2 mole of HCl= $\dfrac{1}{2}\times 0.2=0.1$ mol of $CaC{{l}_{2}}$ .
Therefore, the number of moles of $CaC{{l}_{2}}$ formed in the reaction is 0.1 moles.
So, the correct answer is “Option A”.

Note: The reactant which is completely consumed in a reaction is the limiting reagent. In the given reaction, when the total number of moles of $CaC{{O}_{3}}$, i.e. 0.1 mole has reacted with HCl. No more formation of $CaC{{l}_{2}}$ is possible because there are no moles of $CaC{{O}_{3}}$ left for HCl to react with.