Q.Calcium carbonate reacts with aqueous HCl to give $CaCl_2$ and $CO_2$ according to the reaction given below: $CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)$ What mass of $CaCl_2$ will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of $CaCO_3$? Name the limiting reagent. Calculate the number of moles of $CaCl_2$ formed in the reaction.

Concept: Stoichiometry and Limiting Reagent Determination.

1. Calculate initial moles of reactants:

   Moles of $HCl = \text{Molarity} \times \text{Volume} = 0.76 \text{ M} \times 0.250 \text{ L} = 0.19 \text{ mol}$.

   Molar mass of $CaCO_3 = 40.08 (\text{Ca}) + 12.01 (\text{C}) + 3 \times 16.00 (\text{O}) = 100.09 \text{ g/mol}$.

   Moles of $CaCO_3 = \text{Mass} / \text{Molar mass} = 1000 \text{ g} / 100.09 \text{ g/mol} \approx 9.99 \text{ mol}$.

2. Identify the limiting reagent:

   From the balanced equation, $1 \text{ mol } CaCO_3$ reacts with $2 \text{ mol } HCl$.

   To react completely with $0.19 \text{ mol } HCl$, $0.19/2 = 0.095 \text{ mol } CaCO_3$ is required.

   Since $9.99 \text{ mol } CaCO_3$ is available (which is much more than $0.095 \text{ mol}$), $HCl$ is the limiting reagent.

3. Calculate moles of $CaCl_2$ formed:

   Based on the limiting reagent ($HCl$), the stoichiometry shows $2 \text{ mol } HCl$ produces $1 \text{ mol } CaCl_2$.

   Therefore, $0.19 \text{ mol } HCl$ will produce $0.19/2 = 0.095 \text{ mol } CaCl_2$.

4. Calculate mass of $CaCl_2$ formed: …

Chemical reactions involve specific ratios of reactants combining to form products. Stoichiometry is the branch of chemistry that deals with these quantitative relationships. When we mix reactants, it's rare that they are present in exactly the perfect stoichiometric ratio. Often, one reactant will run out before the other, stopping the reaction. This reactant is called the limiting reagent (or limiting reactant), because it limits the amount of product that can be formed. The other reactant, present in excess, is called the excess reagent.

To solve this problem, we need to:

1. Ensure the chemical equation is balanced, as this provides the crucial mole ratios.
2. Calculate the initial moles of each reactant.
3. Determine which reactant is the limiting reagent by comparing the available moles to the stoichiometric requirements.
4. Use the moles of the limiting reagent to calculate the moles of the product ($CaCl_2$) formed.
5. Convert the moles of $CaCl_2$ to its mass.

Let's work through these steps.

1. Verify the balanced chemical equation.

   The given reaction is:

   $CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)$

   Let's check the atom count on both sides:

   • Ca: 1 on left, 1 on right (Balanced)
   • C: 1 on left, 1 on right (Balanced)
   • O: 3 on left, $2 (in CO_2) + 1 (in H_2O) = 3$ on right (Balanced)
   • H: 2 on left (from $2HCl$), 2 on right (from $H_2O$) (Balanced)
   • Cl: 2 on left (from $2HCl$), 2 on right (from $CaCl_2$) (Balanced)

   The equation is already balanced as provided. This means 1 mole of $CaCO_3$ reacts with 2 moles of $HCl$ to produce 1 mole of $CaCl_2$, 1 mole of $CO_2$, and 1 mole of $H_2O$.

2. Calculate the initial moles of each reactant.

   To do this, we first need the molar masses of the reactants.

   • Molar mass of $CaCO_3$:

     $Ca: 40.08 \text{ g/mol}$

     $C: 12.01 \text{ g/mol}$

     $O: 16.00 \text{ g/mol}$

     Molar mass of $CaCO_3 = 40.08 + 12.01 + (3 \times 16.00) = 100.09 \text{ g/mol}$

   • Molar mass of $HCl$:

     $H: 1.01 \text{ g/mol}$

     $Cl: 35.45 \text{ g/mol}$

     Molar mass of $HCl = 1.01 + 35.45 = 36.46 \text{ g/mol}$

   Now, calculate the moles:

   • Moles of $CaCO_3$: Given mass of $CaCO_3 = 1000 \text{ g}$

*   **Moles of $HCl$:**
    Given volume of $HCl = 250 \text{ mL} = 0.250 \text{ L}$
    Given molarity of $HCl = 0.76 \text{ M}$

3. Identify the limiting reagent.

   The balanced equation tells us that $1 \text{ mol } CaCO_3$ reacts with $2 \text{ mol } HCl$.

   We have $9.990 \text{ mol } CaCO_3$ and $0.19 \text{ mol } HCl$.

   Let's see how much of one reactant is needed to react completely with the other:

   • If all $0.19 \text{ mol}$ of $HCl$ were to react, it would require:

    We have $9.990 \text{ mol } CaCO_3$ available, which is much more than the $0.095 \text{ mol}$ required. This means $CaCO_3$ is in excess.

*   If all $9.990 \text{ mol}$ of $CaCO_3$ were to react, it would require:

    We only have $0.19 \text{ mol } HCl$ available, which is far less than the $19.98 \text{ mol}$ required.

Since $HCl$ is the reactant that will run out first, **$HCl$ is the limiting reagent**. …

Concept: Limiting Reagent & Stoichiometric Calculations

Method: Limiting Reagent Method

This method identifies which reactant gets consumed first, thereby determining the maximum amount of product formed.

Step-by-Step Solution

Step 1: Calculate moles of each reactant

For HCl:

• Volume = 250 mL = 0.250 L
• Molarity = 0.76 M

For CaCO₃:

• Mass given = 1000 g
• Molar mass of CaCO₃ = 40 + 12 + 48 = 100 g/mol

Step 2: Determine the limiting reagent

From the balanced equation:

The mole ratio is: 1 mol CaCO₃ : 2 mol HCl

Check for HCl:

• Moles of HCl needed to react with 10 mol CaCO₃ = $10 \times 2 = 20$ mol
• But we only have 0.19 mol HCl → HCl is insufficient

Check for CaCO₃:

• Moles of CaCO₃ needed to react with 0.19 mol HCl = $\frac{0.19}{2} = 0.095$ mol
• We have 10 mol CaCO₃ → CaCO₃ is in excess

∴ Limiting reagent = HCl

--- …

Common Mistakes Students Make on This Problem

1. Forgetting to Convert Volume to Liters

• The Mistake: Students plug in 250 mL directly into the molarity formula without converting to liters.
• Why It's Wrong: Molarity ($M$) is moles per liter, so $0.76 \, M$ means $0.76$ moles per 1000 mL. Using 250 mL as if it were liters gives an answer that is 4 times too large.
• How to Avoid: Always write the conversion: $250 \, \text{mL} = 0.250 \, \text{L}$ Then use: $\text{moles of HCl} = M \times V = 0.76 \times 0.250$

2. Using the Wrong Mole Ratio for Limiting Reagent

• The Mistake: Comparing moles of HCl directly to moles of $CaCO_3$ without considering the stoichiometric coefficients.
• Why It's Wrong: The balanced equation shows: $1 \, CaCO_3 : 2 \, HCl$ So 1 mole of $CaCO_3$ needs 2 moles of HCl, not 1.
• How to Avoid: Always write the ratio explicitly: $\text{Required moles of HCl} = 2 \times \text{moles of } CaCO_3$ Then compare with available moles of HCl.

3. Miscalculating Molar Mass of $CaCO_3$

• The Mistake: Using $100 \, \text{g/mol}$ as a rounded value when exact calculation is needed, or forgetting to add all atoms.
• Why It's Wrong: $CaCO_3$ = $40 + 12 + (3 \times 16) = 100 \, \text{g/mol}$ — this is correct here, but students often mis-add for other compounds.
• How to Avoid: Write out each element's contribution: $M(CaCO_3) = 40.08 + 12.01 + 48.00 \approx 100.09 \, \text{g/mol}$ (Use atomic masses as given in your exam — usually 40, 12, 16 for simplicity.)

4. Confusing Mass of $CaCO_3$ with Moles

• The Mistake: Treating 1000 g as if it were 1000 moles.
• Why It's Wrong: Mass and moles are different quantities. $\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1000}{100} = 10 \, \text{moles}$
• How to Avoid: Always write the formula: $n = \frac{m}{M}$ and check if your answer is reasonable — 1000 g of a solid cannot be just a few moles if molar mass is ~100.

5. Stopping After Finding the Limiting Reagent

• The Mistake: Identifying the limiting reagent but not using it to calculate the product.
• Why It's Wrong: The limiting reagent determines how much product forms. Using the excess reagent gives a wrong (larger) mass of $CaCl_2$.
• How to Avoid: After finding the limiting reagent, always ask: "Which reactant runs out first?" …