Chemistry Q51 – JEE-MAIN 2025

Home Chemistry Basic ChemistryQuestion #65

C a C O_{3} (s) + 2 H C l (a q) \to C a C l_{2} (a q) + C O_{2} (g) + H_{2} O (l)

Consider the above reaction, what mass of

C a C l_{2}

will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of

C a C O_{3}

? (Given : Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g mol

^{- 1}

, respectively)

Clue 1 of 3: Stoichiometry Basics

Identify the given quantities for each reactant and the product you need to calculate.

🥷

Ninja Strategy•Order of Magnitude Estimation

Quickly estimate the moles of the limiting reactant and the product, then multiply by the approximate molar mass of the product to find the expected order of magnitude for the answer. This helps eliminate options that are too small or too large.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Calculate Moles of Reactants✦ Active

First, determine the molar masses of $C a C O_{3}$ and $H C l$ . Then, calculate the initial moles of each reactant using the given mass and concentration.

Molar mass of C a C O_{3} = 40 + 12 + (3 \times 16) = 100 g/mol

Moles of C a C O_{3} = \frac{1000 g}{100 g/mol} = 10 mol

Molar mass of H C l = 1 + 35.5 = 36.5 g/mol

Moles of H C l = Molarity \times Volume (L) = 0.76 mol/L \times 0.250 L = 0.19 mol

Step 2: Identify Limiting Reactant○ Expand

Compare the mole ratio of reactants to the stoichiometric ratio from the balanced equation ( $1 mol C a C O_{3} : 2 mol H C l$ ) to find the limiting reactant.

From the equation: 1 mol C a C O_{3} reacts with 2 mol H C l

For C a C O_{3} : 10 mol C a C O_{3} would require 10 \times 2 = 20 mol H C l

For H C l : 0.19 mol H C l would require \frac{0.19}{2} = 0.095 mol C a C O_{3}

Since only 0.19 mol H C l is available, which is less than 20 mol, H C l is the limiting reactant.

💡 Teacher's Secret Hint

Always compare the available moles to the required moles based on stoichiometry to correctly identify the limiting reactant.

Step 3: Calculate Mass of Product○ Expand

Use the moles of the limiting reactant ( $H C l$ ) and the stoichiometric ratio to find the moles of $C a C l_{2}$ formed, then convert to mass using its molar mass.

Molar mass of C a C l_{2} = 40 + (2 \times 35.5) = 40 + 71 = 111 g/mol

From the equation: 2 mol H C l produces 1 mol C a C l_{2}

Moles of C a C l_{2} = 0.19 mol H C l \times \frac{1 mol C a C l_{2}}{2 mol H C l} = 0.095 mol C a C l_{2}

Mass of C a C l_{2} = 0.095 mol \times 111 g/mol = 10.545 g

💡 Teacher's Secret Hint

Ensure you use the correct molar mass for the product ( $C a C l_{2}$ ) in the final calculation.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Chemistry Question 51 – JEE-MAIN 2025

Clue 1 of 3: Stoichiometry Basics

🧠 Full Solution Path