A) The limiting reagent is calcium metal.

B) 33.3 g of CaCl₂ is produced.

C) 7.84 L of H₂ gas is evolved.

D) 0.35 mol of H₂ gas is evolved.

Explanation

First, let us write the balanced chemical equation for the reaction between Calcium and Hydrochloric acid:

\[ \text{Ca} + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2 \]

Step 1: Calculate the moles of Calcium (Ca) used.

\[ \text{Moles of Ca} = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{14.0}{40} = 0.35 \text{ mol} \]

Step 2: Check Limiting Reagent. Since HCl is mentioned as "excess", Calcium is the limiting reagent. Statement (A) is correct.

Step 3: Calculate moles of H₂ evolved. From the stoichiometry (1:1 ratio between Ca and H₂):

\[ \text{Moles of H}_2 = 0.35 \text{ mol} \]

Statement (D) is correct.

Step 4: Calculate the volume of H₂ at 1.0 atm and 273 K (STP conditions). At STP, 1 mole of gas occupies 22.4 L.

\[ \text{Volume of H}_2 = 0.35 \times 22.4 = 7.84 \text{ L} \]

Statement (C) is correct.

Step 5: Calculate the mass of CaCl₂ produced. Molar mass of CaCl₂ = 40 + (2 × 35.5) = 40 + 71 = 111 g/mol.

\[ \text{Mass of CaCl}_2 = \text{Moles} \times \text{Molar Mass} = 0.35 \times 111 = 38.85 \text{ g} \]

Statement (B) says 33.3 g is produced, which is incorrect.

Correct Answer: B (Incorrect statement)

Related Theory

The "Mole Concept" and "Stoichiometry" are the bedrock of physical chemistry for competitive exams like JEE Main, JEE Advanced, and NEET. These topics deal with the quantitative relationship between reactants and products in a chemical reaction. Understanding these concepts requires a clear grasp of atomic masses, molar volumes, and the laws of chemical combination.

1. The Mole Concept
A mole is the amount of substance that contains as many entities (atoms, molecules, or ions) as there are atoms in exactly 12 grams of Carbon-12. This number is known as Avogadro's number (\(6.022 \times 10^{23}\)). For any element, the molar mass in grams is numerically equal to its atomic mass in u (unified mass units). In this problem, we converted the mass of Calcium to moles, which is the standard first step in almost all stoichiometric calculations.

2. Stoichiometry of Reactions
Stoichiometry allows us to predict the amount of product formed from a given amount of reactant. The coefficients in a balanced equation represent the molar ratios. For example, in the reaction \(\text{Ca} + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\), the ratio of \(\text{Ca}\) to \(\text{H}_2\) is \(1:1\). This means \(0.35\) moles of Calcium will produce exactly \(0.35\) moles of Hydrogen gas, provided the reaction goes to completion.

3. Limiting and Excess Reagents
The limiting reagent is the reactant that is completely consumed first in a reaction. It "limits" the amount of product that can be formed. Any other reactant present in a quantity greater than required to react with the limiting reagent is called the excess reagent. In the given question, the problem explicitly states that HCl is in excess. This simplifies the problem because we don't have to perform a comparison; we immediately know the yield depends solely on the amount of Calcium provided.

4. Molar Volume of Gases
Under standard conditions of temperature and pressure (STP), the volume of 1 mole of an ideal gas is a constant value. While there are slight variations in definition (IUPAC vs. older standards), for most JEE Main problems, STP is considered as \(273.15 \text{ K}\) and \(1 \text{ atm}\), where the molar volume is \(22.4 \text{ L}\). In this problem, the conditions provided are \(273 \text{ K}\) and \(1.0 \text{ atm}\), matching this standard perfectly. If the conditions were different, we would need to use the Ideal Gas Equation: \(PV = nRT\).

5. Common Pitfalls for JEE Students
- Unbalanced Equations: Students often jump to calculations without balancing the equation. In this case, the \(1:2\) ratio between Ca and HCl is crucial if we were calculating the amount of HCl consumed.
- Atomic Mass Errors: Forgetting that Chlorine is diatomic in its gaseous state (\(\text{Cl}_2\)) but has an atomic mass of \(35.5\) is common. In \(\text{CaCl}_2\), the calculation must include two chlorine atoms.
- STP Conditions: Mixing up \(22.4 \text{ L}\) (atm) and \(22.7 \text{ L}\) (bar) can lead to small errors, though usually, options are far enough apart to distinguish the correct one.
- Reading the Question: The most common mistake is failing to see the word "incorrect". Many students find statement (A) is true and mark it as the answer immediately.

6. Shortcut Tricks
To save time during the exam, instead of calculating everything, look for logical links. For instance, once you find the moles of Calcium (\(0.35\)), you can quickly check if \(0.35 \times 111\) (molar mass of \(\text{CaCl}_2\)) could possibly be \(33.3\). Since \(0.3 \times 111\) is already \(33.3\), then \(0.35 \times 111\) must be higher. This allows you to identify statement (B) as the incorrect one without finishing the exact multiplication.

7. Exam Relevance
This specific question type—where multiple statements must be evaluated—is a favorite in JEE Main. It tests several sub-topics (moles, gas volume, stoichiometry, and limiting reagents) in a single problem. Mastery over these basics is essential for moving on to more complex topics like Redox Titrations, Thermodynamics, and Chemical Equilibrium.

Related JEE Main Questions