Question

Consider the following reaction:
\[ \text{Ca} + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2 \]
\text{We have 14 g Ca reacts with excess of HCl. Choose the incorrect option.}

• A. Mass of CaCl₂ produced is 38.85 g
• B. Mole of H₂ produced is 0.35 mol
• C. Volume of H₂ at STP is 7.84 L
• D. Mass of CaCl₂ produced is 3.885 g

Hint:

When reacting a metal with an acid, the volume of hydrogen gas produced can be calculated from stoichiometry, and the mass of the salt produced depends on the molar mass of the salt.

Answer 1

The Correct Option is D

Step 1: Reaction Stoichiometry.
- Molar mass of calcium (Ca) = 40 g/mol.
- Moles of Ca = \(\frac{14}{40} = 0.35\) mol.
From the balanced reaction, 1 mole of Ca produces 1 mole of CaCl₂. Therefore, moles of CaCl₂ = 0.35 mol.
Step 2: Mass of CaCl₂.
The molar mass of CaCl₂ = \(40 + 2(35.5) = 111 \, \text{g/mol}\).
Mass of CaCl₂ = \(0.35 \times 111 = 38.85 \, \text{g}\).
Step 3: Volume of H₂.
From the balanced equation, 1 mole of H₂ is produced for every mole of Ca. Thus, moles of H₂ = 0.35 mol.
At STP, 1 mole of gas occupies 22.4 L, so volume of H₂ = \(0.35 \times 22.4 = 7.84 \, \text{L}\).
Step 4: Conclusion.
The correct mass of CaCl₂ produced is 38.85 g, so the incorrect option is (4) Mass of CaCl₂ produced is 3.885 g.