For the given reaction; CaCO_3 + 2HCl rightarrow CaCl_2 + H_2O + CO_2 If 90text g CaCO_3 is added to 300text mL of HCl which contains 38.55\%text HCl by mass and has density 1.13text g mL^-1 then which of the following option is correct? Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and 16text g mol^-1 respectively.

Solution & Explanation

### Core Logic Step 1: Calculate moles of HCl available. Density of solution (d) = 1.13text g/mL. Volume (V) = 300text mL. Mass of solution = 300 times 1.13 = 339text g. Mass of pure HCl = 339 times frac38.55100 = 130.68text g. Molar mass of HCl = 36.5text g/mol. Moles of HCl initially = frac130.6836.5 = 3.58text moles. Step 2: Calculate moles of CaCO_3 available. Molar mass of CaCO_3 = 100text g/mol. Moles of CaCO_3 = frac90100 = 0.90text mole. ### Step 1: Determine Limiting Reagent and Reaction Reaction: CaCO_3 + 2HCl rightarrow CaCl_2 + H_2O + CO_2 From the stoichiometry, 1 mole of CaCO_3 requires 2 moles of HCl. 0.90text mole of CaCO_3 requires 0.90 times 2 = 1.80text moles of HCl. Since 3.58 moles of HCl are available, CaCO_3 is the limiting reagent (LR) and HCl is in excess. Moles of HCl remained unreacted = 3.58 - 1.80 = 1.78text moles. Mass of HCl remained = 1.78 times 36.5 = 64.97text g. ### Pattern Recognition Always convert volume, density, and mass percentage into pure mass. Identify Limiting Reagent before finding the leftover. ### 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
Q57 jee_main_2026_21_jan_morning Stoichiometry and Limiting Reagent
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]
  • A. text0.35 mol of H_2text gas is evolved.
  • B. 7.84 mathrm~Ltext of mathrmH_2text gas is evolved.
  • C. text33.3 g of mathrmCaCl_2text is produced.
  • D. textThe limiting reagent is calcium metal.

Solution

### 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
Q55 jee_main_2026_21_jan_evening Percentage Composition and Stoichiometry
By usual analysis, 1.00g of compound (X) gave 1.79g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is: (nearest integer) (Given, molar mass in textg mol^-1: textO = 16, textMg = 24, textP = 31)
  • A. (1) \ 50
  • B. (2) \ 30
  • C. (3) \ 20
  • D. (4) \ 40

Solution

### Related Formula \% text of textP = fractextMoles of textMg_2textP_2textO_7 times 2 times 31textMass of compound times 100 ### Core Logic Molar mass of textMg_2textP_2textO_7 = 2(24) + 2(31) + 7(16) = 48 + 62 + 112 = 222 text g/mol. textPercentage of P = fracleft(frac1.79222 times 2 times 31right)1 times 100 = 49.99\% approx 50\% ### Step 1: Final Calculation Rounding to the nearest integer gives 50. ### Pattern Recognition Sees: Gravimetric analysis involving magnesium pyrophosphate. Trap: Forgetting the factor of 2 for phosphorus atoms in textMg_2textP_2textO_7. ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q56 jee_main_2026_22_january_morning Stoichiometry and Molar Volume
In the reaction, 2Al(s) + 6HCl(aq) rightarrow 2Al^3+(aq) + 6Cl^-(aq) + 3H_2(g)
  • A. text11.2 L H_2(g) text at STP is produced for every mole of HCl consumed.
  • B. text67.2 L H_2(g) text at STP is produced for every mole of Al that reacts.
  • C. text12 L HCl(aq) is consumed for every 6L H_2(g) text produced.
  • D. text33.6 L H_2(g) text is produced regardless of temperature and pressure for every mole of Al that reacts.

Solution

### Related Formula textVolume of gas at STP = textMoles times 22.4 text L ### Core Logic From the balanced chemical equation: 2Al(s) + 6HCl(aq) rightarrow 2Al^3+(aq) + 6Cl^-(aq) + 3H_2(g) 6 moles of HCl produce 3 moles of H_2. Therefore, 1 mole of HCl produces frac36 = 0.5 moles of H_2. Volume of H_2 produced at STP for 1 mole of HCl: V = 0.5 times 22.4text L = 11.2text L ### Step 1: Check other options Option (2): 2 moles of Al produce 3 moles of H_2. 1 mole Al produces 1.5 moles H_2 = 1.5 times 22.4 = 33.6text L (Incorrect). Option (4) is incorrect because volume depends on temperature and pressure. ### Pattern Recognition Standard stoichiometry. Directly map the mole ratio from the balanced equation to molar volume at standard conditions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q53 jee_main_2026_22_january_evening Limiting Reagent and Stoichiometry
textA + 2textB rightarrow textAB_2 36.0text g of 'A' (Molar mass: 60text g mol^-1) and 56.0text g of 'B' (Molar mass: 80text g mol^-1) are allowed to react. Which of the following statements are correct? (A) 'A' is the limiting reagent (B) 77.0text g of textAB_2 is formed (C) Molar mass of textAB_2 is 140text g mol^-1 (D) 15.0text g of A is left unreacted after the completion of reaction. Choose the correct answer from the options given below:
  • A. C and D only
  • B. A and C only
  • C. B and D only
  • D. A and B only

Solution

### Related Formula textMoles (n) = fractextGiven MasstextMolar Mass textMolar Mass of textAB_2 = M_A + 2 M_B ### Core Logic Step 1: Calculate initial moles: n_A = frac3660 = 0.6text mol n_B = frac5680 = 0.7text mol Step 2: Identify Limiting Reagent (LR): textRatio for A = frac0.61 = 0.6, quad textRatio for B = frac0.72 = 0.35 Since ratio of B is smaller, B is the limiting reagent. Step 3: Evaluate product formed and remaining reactant: textMolar mass of textAB_2 = 60 + 2(80) = 220text g mol^-1 textMoles of textAB_2 text formed = 0.35text mol textMass of textAB_2 text formed = 0.35 times 220 = 77.0text g quad text(Statement B is correct) textMoles of A reacted = 0.35text mol textMoles of A remaining = 0.6 - 0.35 = 0.25text mol textMass of A remaining = 0.25 times 60 = 15.0text g quad text(Statement D is correct) ### Pattern Recognition Sees: Initial masses of reactants in stoichiometric equation. Shortcut: Compare n/textcoefficient to find LR, then compute formed product mass and unreacted mass directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

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