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 : Molar mass in mathrmg \ mol^-1 of Ca–40, Cl–35.5, H–1]

Solution & Explanation

### Related Formula mathrmCa_(mathrms) + 2mathrmHCl_(mathrmg) longrightarrow mathrmCaCl_2(mathrms) + mathrmH_2(mathrmg) ### Core Logic Number of moles of Calcium (n_Ca) = fractextGiven masstextMolar mass = frac14.040 = 0.35 mol. Since HCl is in excess, Calcium is the limiting reagent. From stoichiometry, 1 mole of Ca produces 1 mole of H_2 gas and 1 mole of CaCl_2. Moles of H_2 evolved = 0.35 mol. Volume of H_2 at STP (1 atm, 273 K) = 0.35 times 22.4 L = 7.84 L. Mass of CaCl_2 produced = 0.35 times textMolar mass of CaCl_2 = 0.35 times (40 + 71) = 0.35 times 111 = 38.85 g. Option (3) states 33.3 g of CaCl_2 is produced, which is incorrect. ### Step 1: Final Conclusion Statement (3) is incorrect. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

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Q56 jee_main_2026_21_jan_morning Eudiometry
80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :
  • A. mathrmC_2mathrmH_4
  • B. mathrmC_4mathrmH_10
  • C. mathrmC_2mathrmH_2
  • D. mathrmC_2mathrmH_6

Solution

### Related Formula mathrmC_mathrmxmathrmH_mathrmy(g) + left(mathrmx + fracmathrmy4right)mathrmO_2(g) longrightarrow mathrmxCO_2(g) + fracmathrmy2mathrmH_2mathrmO_(ell) ### Core Logic Let the volume of hydrocarbon be V = 80 mL. Initial volume of O_2 = 264 mL. At 273 K, H_2O is liquid, so its volume is neglected. Volume of CO_2 formed = 80x mL. Volume of O_2 used = 80left(x + fracy4right) mL. Unreacted O_2 = 264 - 80left(x + fracy4right) mL. Total residual volume = V_CO_2 + V_unreacted \ O_2 = 224 mL. 80x + 264 - 80left(x + fracy4right) = 224 264 - frac80y4 = 224 40 = 20y implies y = 2 After treatment with KOH, CO_2 is absorbed. The remaining volume is unreacted O_2, which is 64 mL. 264 - 80left(x + fracy4right) = 64 Substitute y = 2: 264 - 80left(x + frac12right) = 64 264 - 80x - 40 = 64 224 - 80x = 64 80x = 160 implies x = 2 The hydrocarbon is mathrmC_2mathrmH_2. ### Pattern Recognition Volume decrease by KOH indicates the volume of CO_2 produced. V_CO_2 = 224 - 64 = 160 mL. V_HC = 80 mL. So x = frac16080 = 2. Total volume reduction = 264 - 64 = 200 mL (O_2 consumed). O_2 consumed = 80(x + y/4) = 200 implies 2 + y/4 = 2.5 implies y/4 = 0.5 implies y = 2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q33 jee_main_2025_02_april_morning Stoichiometry and Limiting Reagent
mathrmCaCO_3(s) + 2HCl(aq) rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l) Consider the above reaction, what mass of mathrmCaCl_2 will be formed if 250mathrm~mL of 0.76mathrm~M HCl reacts with 1000mathrm~g of mathrmCaCO_3? (Given: Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5mathrm~g cdot mol^-1, respectively)
  • A. (1)\ 3.908mathrm~g
  • B. (2)\ 2.636mathrm~g
  • C. (3)\ 10.545mathrm~g
  • D. (4)\ 5.272mathrm~g

Solution

### Related Formula Molarity conversion relation matrix: textMoles = textMolarity (M) times textVolume (L) textMass = textMoles times textMolar Mass ### Core Logic Let's perform molar quantities verification row-by-row: * Molar mass properties: mathrmCaCO_3 = 100mathrm~g/mol, mathrmCaCl_2 = 40 + (35.5 times 2) = 111mathrm~g/mol. * Initial chemical moles calculated: textMoles of mathrmCaCO_3 = frac1000100 = 10mathrm~mol textMoles of mathrmHCl = 0.76 times frac2501000 = 0.19mathrm~mol * Determine the limiting reactant via stoichiometric ratios: mathrmHCl acts as the Limiting Reagent (L.R.) because its proportional structural requirement is much smaller. * Moles of product mathrmCaCl_2 formed based on L.R. configuration: textMoles of mathrmCaCl_2 = frac0.192 = 0.095mathrm~mol textMass of mathrmCaCl_2 = 0.095 times 111 = 10.545mathrm~g ### Pattern Recognition Always compare the available moles divided by the respective stoichiometric coefficients to quickly find the Limiting Reagent: 10/1 gg 0.19/2. This trick saves execution seconds during complex numeric problems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q27 jee_main_2025_03_april_evening Molarity and Stoichiometry of Neutralization
10mathrm~mL of 2mathrm~M~NaOH solution is added to 20mathrm~mL of 1mathrm~M~HCl solution kept in a beaker. Now, 10mathrm~mL of this mixture is poured into a volumetric flask of 100mathrm~mL containing 2 moles of mathrmHCl and made the volume upto the mark with distilled water. The solution in this flask is :
  • A. 0.2mathrm~M~NaCl solution
  • B. 20mathrm~M~HCl solution
  • C. 10mathrm~M~HCl solution
  • D. Neutral solution

Solution

### Related Formula Number of millimoles (n) is given by: n = M times V_mathrmmL Molarity (M) of a diluted mixture: M = fractextTotal molestextTotal Volume in Liters ### Core Logic Evaluate the first mixing step to determine the net acid-base state: - Millimoles of mathrmNaOH = 10mathrm~mL times 2mathrm~M = 20mathrm~mmol - Millimoles of mathrmHCl = 20mathrm~mL times 1mathrm~M = 20mathrm~mmol Since millimoles are equal, mathrmHCl and mathrmNaOH completely neutralize each other, producing a neutral aqueous salt solution. ### Step 1: Analyze transfer to volumetric flask Taking 10mathrm~mL of this neutralized solution provides no excess mathrmH^+ or mathrmOH^- ions. It is added to a volumetric flask containing 2mathrm~mol of pure mathrmHCl. ### Step 2: Calculate final molarity of mathrmHCl The volume of the flask is made up to 100mathrm~mL = 0.1mathrm~L: M = frac2mathrm~mol0.1mathrm~L = 20mathrm~M Hence, the resulting solution is 20mathrm~M~HCl. ### Pattern Recognition Stoichiometric neutralizations are evaluated by setting up mole/millimole balance charts. Once stoichiometric equivalence (MV_textacid = MV_textbase) is reached, any sub-aliquot of that solution remains completely neutral. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry Class 11 Chemistry: Ionic Equilibrium
Q37 jee_main_2025_03_april_evening Stoichiometry of Gas Evolution
Mass of magnesium required to produce 220mathrm~mL of hydrogen gas at STP on reaction with excess of dil. HCl is : Given: Molar mass of Mg is 24mathrm~g~mol^-1 .
  • A. 235.7 g
  • B. 0.24 mg
  • C. 236 mg
  • D. 2.444 g

Solution

### Related Formula The balanced chemical equation for the displacement reaction is: mathrmMg(s) + 2mathrmHCl(aq) rightarrow mathrmMgCl_2mathrm(aq) + mathrmH_2mathrm(g) At STP, 1 mole of any ideal gas occupies a volume of 22.4mathrm~L = 22400mathrm~mL. ### Core Logic From the stoichiometry of the reaction: - 1 mole of mathrmMg (24mathrm~g) produces 1 mole of mathrmH_2 (22400mathrm~mL at STP). ### Step 1: Calculate moles of mathrmH_2 gas produced n_mathrmH_2 = frac220mathrm~mL22400mathrm~mL/mol approx 9.8214 times 10^-3mathrm~mol ### Step 2: Calculate mass of Magnesium required Since the molar ratio of \mathrm{Mg} to \mathrm{H}_2 is 1:1: n_mathrmMg = 9.8214 times 10^-3mathrm~mol textMass of Mg = 9.8214 times 10^-3mathrm~mol times 24mathrm~g/mol textMass of Mg approx 0.2357mathrm~g = 235.7mathrm~mg approx 236mathrm~mg This matches Option (3). ### Pattern Recognition Always keep a close eye on unit prefixes in options. A mass of 0.2357\mathrm{~g} corresponds to 235.7\mathrm{~mg}, which rounds directly to 236\mathrm{~mg}, whereas 235.7\mathrm{~g}$ is off by a factor of 1000. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

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