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

Let's analyze the given chemical reaction and determine the correct mass of \(\text{CaCl}_2\) produced.

The balanced chemical reaction is:

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

1. First, calculate the moles of calcium (\( \text{Ca} \)) reacted:
   • Molar mass of calcium (\( \text{Ca} \)) is approximately 40 g/mol.
   • Moles of \(\text{Ca}\) = \(\frac{14 \text{ g}}{40 \text{ g/mol}} = 0.35 \text{ mol}\)
2. From the balanced reaction, 1 mole of \(\text{Ca}\) produces 1 mole of \(\text{CaCl}_2\).
3. Therefore, moles of \(\text{CaCl}_2\) produced = moles of \(\text{Ca}\) = 0.35 mol.
4. Calculate the mass of \(\text{CaCl}_2\):
   • Molar mass of \(\text{CaCl}_2\) is \((40 + 2 \times 35.5) \text{ g/mol} = 111 \text{ g/mol}\).
   • Mass of \(\text{CaCl}_2\) = \(0.35 \text{ mol} \times 111 \text{ g/mol} = 38.85 \text{ g}\)

Now, let's verify each option against what we calculated:

• Option 1: Mass of \(\text{CaCl}_2\) produced is 38.85 g. This is correct based on our calculation.
• Option 2: Mole of \(\text{H}_2\) produced is 0.35 mol. Correct, because 0.35 mol of \(\text{Ca}\) produces 0.35 mol of \(\text{H}_2\).
• Option 3: Volume of \(\text{H}_2\) at STP is 7.84 L.
  • At STP, 1 mole of gas occupies 22.4 L.
  • Volume of \(\text{H}_2\) = \(0.35 \text{ mol} \times 22.4 \text{ L/mol} = 7.84 \text{ L}\).
  • This is correct.
• Option 4: Mass of \(\text{CaCl}_2\) produced is 3.885 g. This is incorrect, as the correct mass is 38.85 g.

The incorrect option, as determined from the calculations above, is the one stating that the mass of \(\text{CaCl}_2\) produced is 3.885 g.