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JEE Chemistry 2025 Question with Solution

Consider the above reaction, what mass of CaCl2\text{CaCl}_2 will be formed if 250 mL250 \, \text{mL} of 0.76 M0.76 \, \text{M} HCl\text{HCl} reacts with 1000 g1000 \, \text{g} of CaCO3\text{CaCO}_3?

  • A

    3.908 g3.908 \, \text{g}

  • B

    2.636 g2.636 \, \text{g}

  • C

    10.545 g10.545 \, \text{g}

  • D

    5.272 g5.272 \, \text{g}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: 250 mL250 \, \text{mL} of 0.76 M0.76 \, \text{M} HCl\text{HCl} reacts with 1000 g1000 \, \text{g} of CaCO3\text{CaCO}_3.

Find: Mass of CaCl2\text{CaCl}_2 formed.

The balanced chemical equation is

CaCO3(s)+2HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)

Molar masses used:

Molar mass of CaCO3=40+12+3(16)=100 g/mol\text{Molar mass of } \text{CaCO}_3 = 40 + 12 + 3(16) = 100 \, \text{g/mol} Molar mass of CaCl2=40+2(35.5)=111 g/mol\text{Molar mass of } \text{CaCl}_2 = 40 + 2(35.5) = 111 \, \text{g/mol}

Moles of HCl\text{HCl}:

Volume=250 mL=0.250 L\text{Volume} = 250 \, \text{mL} = 0.250 \, \text{L} Moles of HCl=0.76×0.250=0.19 mol\text{Moles of HCl} = 0.76 \times 0.250 = 0.19 \, \text{mol}

Moles of CaCO3\text{CaCO}_3:

Moles of CaCO3=1000100=10 mol\text{Moles of } \text{CaCO}_3 = \frac{1000}{100} = 10 \, \text{mol}

From the stoichiometry, 11 mole of CaCO3\text{CaCO}_3 reacts with 22 moles of HCl\text{HCl}.

Required moles of CaCO3\text{CaCO}_3 for 0.19 mol0.19 \, \text{mol} HCl\text{HCl}:

0.19×12=0.095 mol0.19 \times \frac{1}{2} = 0.095 \, \text{mol}

Since 10 mol10 \, \text{mol} of CaCO3\text{CaCO}_3 is available, CaCO3\text{CaCO}_3 is in excess and HCl\text{HCl} is the limiting reactant.

Moles of CaCl2\text{CaCl}_2 formed:

Moles of CaCl2=0.19×12=0.095 mol\text{Moles of } \text{CaCl}_2 = 0.19 \times \frac{1}{2} = 0.095 \, \text{mol}

Mass of CaCl2\text{CaCl}_2 formed:

Mass of CaCl2=0.095×111=10.545 g\text{Mass of } \text{CaCl}_2 = 0.095 \times 111 = 10.545 \, \text{g}

Therefore, the mass of CaCl2\text{CaCl}_2 formed is 10.545 g10.545 \, \text{g}. The correct option is C.

Limiting Reactant Shortcut

Given: 0.19 mol0.19 \, \text{mol} of HCl\text{HCl} and excess CaCO3\text{CaCO}_3.

Find: Mass of CaCl2\text{CaCl}_2 formed.

Because the reaction is

CaCO3+2HCl→CaCl2+CO2+H2O\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}

22 moles of HCl\text{HCl} give 11 mole of CaCl2\text{CaCl}_2.

So directly,

Moles of CaCl2=0.192=0.095 mol\text{Moles of } \text{CaCl}_2 = \frac{0.19}{2} = 0.095 \, \text{mol}

Now convert moles to mass using 111 g/mol111 \, \text{g/mol}:

Mass=0.095×111=10.545 g\text{Mass} = 0.095 \times 111 = 10.545 \, \text{g}

This shortcut works because CaCO3\text{CaCO}_3 is present in a very large excess compared to the amount needed for 0.19 mol0.19 \, \text{mol} of HCl\text{HCl}. Therefore, the correct option is C.

Common mistakes

  • A common mistake is treating CaCO3\text{CaCO}_3 as the limiting reactant because its mass looks important. This is wrong because limiting reactant depends on moles, not directly on mass. First convert each reactant to moles, then compare using the balanced equation.

  • Students may forget to convert 250 mL250 \, \text{mL} to 0.250 L0.250 \, \text{L} before using molarity. This is wrong because molarity is in mol/L\text{mol/L}. Always convert volume into liters before applying moles=M×V\text{moles} = M \times V.

  • Another mistake is using a 1:11:1 mole ratio between HCl\text{HCl} and CaCl2\text{CaCl}_2. This is wrong because the balanced equation shows 2HCl:1CaCl22\text{HCl} : 1\text{CaCl}_2. Use the stoichiometric coefficients exactly as written in the balanced reaction.

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