Question

14.0 g of calcium metal is allowed to react with excess HCl at 1.0 atm pressure and 273 K. Which of the following statements is incorrect?
\([Given: \text{Molar mass of } Ca = 40, Cl = 35.5, H = 1 \text{ g mol}^{-1}] \)

• A. 0.35 mol of \(H_{2}\) gas is evolved.
• B. 7.84 L of \(H_{2}\) gas is evolved.
• C. The limiting reagent is calcium metal.
• D. 33.3 g of \(CaCl_{2}\) is produced.

Hint:

Always identify the limiting reagent first in stoichiometry problems. The amount of product formed depends solely on the limiting reagent.

Answer 1

The Correct Option is D

   Step 1: Reaction Stoichiometry
The balanced chemical equation is: \[ Ca(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2(g) \] Step 2: Calculate Moles of Calcium
\[ \text{Moles of Ca} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{14.0}{40} = 0.35 \text{ mol} \] Since HCl is in excess, Calcium is the limiting reagent. (Statement 3 is Correct). Step 3: Analyze Products
\begin{itemize} \item Moles of \(H_2\): According to stoichiometry, 1 mol Ca produces 1 mol \(H_2\). \[ n_{H_2} = 0.35 \text{ mol} \] (Statement 1 is Correct). \item Volume of \(H_2\) at STP: Given \(P = 1\) atm and \(T = 273\) K (STP conditions). \[ V = n \times 22.4 \text{ L} = 0.35 \times 22.4 = 7.84 \text{ L} \] (Statement 2 is Correct). \item Mass of \(CaCl_2\): Moles of \(CaCl_2\) produced = 0.35 mol. Molar mass of \(CaCl_2 = 40 + 2(35.5) = 40 + 71 = 111 \text{ g/mol}\). \[ \text{Mass} = 0.35 \times 111 = 38.85 \text{ g} \] Statement 4 claims 33.3 g is produced, which is Incorrect. \end{itemize} Final Answer: The incorrect statement is option 4.