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: Calculation of Moles of Calcium

\[ \text{Moles of Ca} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{14.0}{40} = 0.35 \text{ mol} \] Since hydrochloric acid is present in excess, calcium is the limiting reagent.
Hence, Statement 3 is correct.

Step 3: Analysis of Products

(i) Moles of Hydrogen Gas Produced
From the balanced equation, 1 mol of calcium produces 1 mol of hydrogen gas.
\[ n_{H_2} = 0.35 \text{ mol} \] Therefore, Statement 1 is correct.

(ii) Volume of Hydrogen Gas at STP
At STP, 1 mol of gas occupies 22.4 L.
\[ V = 0.35 \times 22.4 = 7.84 \text{ L} \] Hence, Statement 2 is correct.

(iii) Mass of Calcium Chloride Formed
Moles of \(CaCl_2\) formed = 0.35 mol.

Molar mass of \(CaCl_2\): \[ = 40 + 2(35.5) = 111 \text{ g/mol} \] Mass formed: \[ = 0.35 \times 111 = 38.85 \text{ g} \] Statement 4 claims that 33.3 g of \(CaCl_2\) is produced, which is incorrect.

Final Answer:
The incorrect statement is Option 4.