Introduction to Limiting Reactants
In chemical reactions, the limiting reactant is the substance that is completely consumed first, thereby determining the maximum amount of product that can be formed. Understanding this concept is crucial for stoichiometry.
The Balanced Equation
First, we must write and balance the reaction between Calcium Carbonate ($CaCO_3$) and Hydrochloric acid ($HCl$):
$$CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$$
From the equation, 1 mole of $CaCO_3$ reacts with 2 moles of $HCl$.
Step-by-Step Solution
Step 1: Identify the limiting reactant
Calculate moles of $CaCO_3$: Molar Mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \text{ g/mol}$. Moles of $CaCO_3 = \frac{25 \text{ g}}{100 \text{ g/mol}} = 0.25 \text{ moles}$.
Compare with $HCl$: We have $0.75 \text{ moles}$ of $HCl$. According to the ratio (1:2), $0.25 \text{ moles}$ of $CaCO_3$ requires $0.25 \times 2 = 0.50 \text{ moles}$ of $HCl$.
Since we have $0.75 \text{ moles}$ of $HCl$ (which is more than the $0.50 \text{ moles}$ required), $CaCO_3$ is the limiting reactant.
Step 2: Calculate mass of $CaCl_2$
From the stoichiometry, 1 mole of $CaCO_3$ produces 1 mole of $CaCl_2$. So, $0.25 \text{ moles}$ of $CaCO_3$ produces $0.25 \text{ moles}$ of $CaCl_2$.
Molar Mass of $CaCl_2 = 40 + (2 \times 35.5) = 111 \text{ g/mol}$. Mass of $CaCl_2 = 0.25 \text{ mol} \times 111 \text{ g/mol} = 27.75 \text{ g}$.
Step 3: Calculate number of water molecules
From the stoichiometry, 1 mole of $CaCO_3$ produces 1 mole of $H_2O$. So, $0.25 \text{ moles}$ of $H_2O$ are produced.
Number of molecules = Moles $\times$ Avogadro's number ($N_A = 6.022 \times 10^{23}$) Number of molecules = $0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23} \text{ molecules}$.