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Consider the following reaction: $Ca+2HCl \rightarrow CaCl_2+H_2$. If 14g of calcium reacts with excess HCl, choose the incorrect option.
APPLY COMPETENCY 4 Marks
Concept Application
50%
Calculation / Logic
50%
Target Level
MEDIUM
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APPLY COMPETENCY MEDIUM

Q: Consider the following reaction: $Ca+2HCl \rightarrow CaCl_2+H_2$. If 14g of calcium reacts with excess HCl, choose the incorrect option.

Question Analysis & Solution

Step-by-Step Solution

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

    The molar mass of Ca is 40 g/mol.

    Moles of Ca = Mass of Ca / Molar mass of Ca = 14 g / 40 g/mol = 0.35 mol

  2. Step 2: Determine the moles of $CaCl_2$ and $H_2$ produced

    From the balanced equation, $Ca + 2HCl \rightarrow CaCl_2 + H_2$, 1 mole of Ca produces 1 mole of $CaCl_2$ and 1 mole of $H_2$.

    Therefore, 0.35 mol of Ca will produce 0.35 mol of $CaCl_2$ and 0.35 mol of $H_2$.

  3. Step 3: Calculate the mass of $CaCl_2$ produced

    The molar mass of $CaCl_2$ is 40 + 2(35.5) = 40 + 71 = 111 g/mol.

    Mass of $CaCl_2$ = Moles of $CaCl_2$ * Molar mass of $CaCl_2$ = 0.35 mol * 111 g/mol = 38.85 g

  4. Step 4: Calculate the volume of $H_2$ produced at STP

    At STP, 1 mole of any gas occupies 22.4 L.

    Volume of $H_2$ = Moles of $H_2$ * 22.4 L/mol = 0.35 mol * 22.4 L/mol = 7.84 L

  5. Step 5: Analyze the options

    • (A) Mass of $CaCl_2$ produced is 38.85 g - Correct
    • (B) Moles of $H_2$ produced is 0.35 mol - Correct
    • (C) Volume of $H_2$ produced at STP is 7.84 L - Correct
    • (D) Mass of $CaCl_2$ produced is 3.885 g - Incorrect

Correct Answer: D

AI Suggestion: Option D
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