Question

Consider the following reaction: \[ {Ca + 2HCl → CaCl2 + H2} \] If \(14\,\text{g}\) of calcium reacts with excess \( {HCl} \), choose the {incorrect} option.

• A. Mass of \( {CaCl2} \) produced is \(38.85\,\text{g}\)
• B. Moles of \( {H2} \) produced is \(0.35\,\text{mol}\)
• C. Volume of \( {H2} \) produced at STP is \(7.84\,\text{L}\)
• D. Mass of \( {CaCl2} \) produced is \(3.885\,\text{g}\)

Hint:

For stoichiometry problems:
Convert given mass into moles first
Use mole ratios from balanced equation
At STP, \(1\) mole of any gas occupies \(22.4\,\text{L}\)

Answer 1

The Correct Option is D

Step 1: Calculate moles of calcium: \[ \text{Moles of Ca} = \frac{14}{40} = 0.35\,\text{mol} \]
Step 2: From the balanced equation: \[ 1\,\text{mol Ca} \rightarrow 1\,\text{mol CaCl}_2 + 1\,\text{mol H}_2 \] So, \[ \text{Moles of } {CaCl2} = 0.35\,\text{mol} \] \[ \text{Moles of } \{H2} = 0.35\text{mol} \] 
Step 3: Mass of \( {CaCl2} \): \[ \text{Molar mass of } {CaCl2} = 40 + 2(35.5) = 111\,\text{g mol}^{-1} \] \[ \text{Mass} = 0.35 \times 111 = 38.8,\text{g} \] 
Step 4: Volume of hydrogen gas at STP: \[ V = 0.35 \times 22.4 = 7.84\,\text{L} \] 
Step 5: Hence, option (D) is incorrect.