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 :

Solution & Explanation

### 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

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Some Basic Concepts Of Chemistry

More Some Basic Concepts of Chemistry Previous-Year Questions — Page 3

Q47 jee_main_2025_04_april_evening Stoichiometry
The amount of calcium oxide produced on heating 150mathrm~kg limestone (75% pure) is _________________ kg. (Nearest integer) Given : Molar mass (in g mol^-1 ) of Ca-40, O-16, C-12
Numerical Answer. Answer: 62.5 to 63.5

Solution

### Related Formula mathrmCaCO_3 xrightarrowDelta mathrmCaO + mathrmCO_2 textPure mass = textTotal mass times fractextPurity \%100 ### Core Logic 1. Find the pure mass of calcium carbonate (CaCO_3) present: textMass of CaCO_3 = 150 times frac75100 = 112.5 mathrm~kg = 112500 mathrm~g 2. Convert this mass into moles (Molar mass of CaCO_3 = 40+12+48 = 100 mathrm~gcdot mol^-1): textmoles of CaCO_3 = frac112500100 = 1125 text moles 3. From the stoichiometry of the reaction, 1 mole of CaCO_3 yields 1 mole of CaO: textmoles of CaO = 1125 text moles textMass of CaO = 1125 times 56 mathrm~g = 63000 mathrm~g = 63 mathrm~kg ### Pattern Recognition Always multiply by the purity fraction first before entering regular stoichiometric conversion chains. Since the formula weight of limestone is exactly 100, tracking percentages directly mirrors mole factors seamlessly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q48 jee_main_2025_04_april_morning Concentration Terms
Fortification of food with iron is done using mathrmFeSO_4cdot 7H_2O. The mass in grams of the mathrmFeSO_4cdot 7H_2O required to achieve 12mathrm~ppm of iron in 150mathrm~kg of wheat is _______. (Nearest integer) [Given: Molar mass of Fe, S and O respectively are 56, 32 and 16mathrm~g~mol^-1]
Numerical Answer. Answer: 9 to 9

Solution

### Related Formula textppm = fractextMass of solute (g)textTotal mass of solution/mixture (g) times 10^6 ### Core Logic Let the required mass of pure iron be wmathrm~g. The total mass of the wheat mixture is 150mathrm~kg = 150 times 10^3mathrm~g. Applying the parts-per-million concentration condition: 12 = fracw150 times 10^3 times 10^6 implies 12 = w times 6.666 w = frac12 times 150 times 10^310^6 = 1.8mathrm~g text of Iron Now, determine the molar mass of the complete green vitriol salt crystal template, mathrmFeSO_4 cdot 7H_2O: M = 56 + 32 + (4 times 16) + (7 times 18) = 56 + 32 + 64 + 126 = 278mathrm~g~mol^-1 Set up a stoichiometric mass balance proportion to find the total salt mass w_1: textMoles of Fe = frac1.856 = fracw_1278 w_1 = frac1.8 times 27856 = frac500.456 approx 8.935mathrm~g Rounding off to the nearest integer value gives **9**. ### Pattern Recognition Always convert concentration metrics back to absolute molar mass equivalence values before distributing across full hydrated molecular templates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q48 jee_main_2025_07_april_evening Stoichiometry and Limiting Reagent
Butane reacts with oxygen to produce carbon dioxide and water following the equation given below: [cite: 461, 463] mathrmC_4H10(g) + frac132O_2(g) ightarrow 4CO_2(g) + 5H_2O(l) If 174.0text kg of butane is mixed with 320.0text kg of textO_2, the volume of water formed in litres is dots. (Nearest integer) [cite: 465, 466] [Given: (a) Molar mass of textC, textH, textO are 12, 1, 16text g mol^-1 respectively, (b) Density of water = 1text g mL^-1] [cite: 467, 468]
Numerical Answer. Answer: 137.5 to 138.5

Solution

### Related Formula textMoles (n) = fractextMass in gramstextMolar Mass textVolume of water (V) = fractextMass of watertextDensity of water ### Core Logic First, calculate initial molar quantities for both reactants: - Molar mass of textC_4textH_10 = 4(12) + 10(1) = 58text g/mol - Initial Moles of butane = frac174.0 times 10^3text g58text g/mol = 3000text mol = 3 times 10^3text mol - Molar mass of textO_2 = 32text g/mol - Initial Moles of oxygen = frac320.0 times 10^3text g32text g/mol = 10000text mol = 10 times 10^3text mol ### Step 1: Identify Limiting Reagent Let's test the stoichiometric requirements via calculation ratios: - Ratio for textC_4textH_10 = frac30001 = 3000 - Ratio for textO_2 = frac1000013/2 = 1538.46 Since the ratio for textO_2 is lower, oxygen behaves as the **limiting reagent** and commands the output steps. ### Step 2: Compute Water Yield Using stoichiometric proportions defined by the balanced reaction field: textMoles of textH2textO = 5 times fractextMoles of textO213/2 = 5 times frac213 times 10000 = frac10000013text mol Convert moles to mass (MtextH2textO = 18text g/mol): textMass of water = frac10000013 times 18 = 138461.5text g approx 138.46text kg Since density = 1text g/mL = 1text kg/L, the net volume is exactly: Vtextwater = 138.46text Litres approx 138text Litres ### Pattern Recognition Limiting reagent shortcut: Always check ratios (moles / stoichiometric coefficient) right away. Do not spend time calculating theoretical products based on butane before establishing whether oxygen runs out first. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q36 jee_main_2025_24_jan_evening Empirical and Molecular Formulae
The elemental composition of a compound is 54.2\%text C, 9.2\%text H and 36.6\%text O. If the molar mass of the compound is 132text g mol^-1, the molecular formula of the compound is: [Given : The relative atomic mass of mathrmC : mathrmH : mathrmO = 12 : 1 : 16]
  • A. mathrmC_4mathrmH_9mathrmO_3
  • B. mathrmC_6mathrmH_12mathrmO_6
  • C. mathrmC_6mathrmH_12mathrmO_3
  • D. mathrmC_4mathrmH_8mathrmO_2

Solution

### Core Logic Let's find the empirical formula by calculating the relative molar ratios of each element: * Carbon (C): textMoles = frac54.212 = 4.516 * Hydrogen (H): textMoles = frac9.21 = 9.200 * Oxygen (O): textMoles = frac36.616 = 2.287 Next, divide each value by the smallest number of moles (2.287) to get the simplest whole-number ratio: * C: frac4.5162.287 approx 1.97 approx 2 * H: frac9.2002.287 approx 4.02 approx 4 * O: frac2.2872.287 = 1 Thus, the Empirical Formula is mathrmC_2mathrmH_4mathrmO. ### Step 2: Molecular Formula Determination 1. Calculate the empirical formula mass: textEmpirical Mass = (2 cdot 12) + (4 cdot 1) + 16 = 24 + 4 + 16 = 44text g/mol 2. Find the multiplier (n): n = fractextMolar MasstextEmpirical Mass = frac13244 = 3 3. Determine the molecular formula: textMolecular Formula = (mathrmC_2mathrmH_4mathrmO)_3 = mathrmC_6mathrmH_12mathrmO_3 This matches Option (3). ### Pattern Recognition Always determine the empirical formula ratio first. Then, compare the empirical mass with the given molar mass to find the whole-number multiplier (n). Multiplication gives the definitive molecular formula. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q50 jee_main_2025_24_jan_morning Stoichiometry and Limiting Reagent
Consider the following reaction occurring in the blast furnace. mathrm F e _ 3 mathrm O _ 4 (mathrm s) + 4 mathrm C O _ (mathrm g) rightarrow 3 mathrm F e _ (mathrm l) + 4 mathrm C O _ 2 (mathrm g) x kg of iron is produced when 2.32times 10^3mathrmkg mathrmFe_3O_4 and 2.8times 10^2mathrmkg mathrmCO are brought together in the furnace. The value of x is ______ (nearest integer) {Given: Molar mass of mathrmFe_3O_4 = 232mathrmg mol^-1 Molar mass of mathrmCO = 28mathrmg mol^-1 Molar mass of mathrmFe = 56mathrmg mol^-1}
Numerical Answer. Answer: 420 to 420

Solution

### Related Formula textMoles (n) = fractextMass in gramstextMolar Mass ### Core Logic First, calculate the input moles for each reactant: - Moles of Fe_3O_4 = frac2.32 times 10^3 times 10^3text g232text g mol^-1 = 10,000text moles - Moles of CO = frac2.8 times 10^2 times 10^3text g28text g mol^-1 = 10,000text moles Next, identify the limiting reagent by comparing the available moles to the stoichiometric coefficients: - For Fe_3O_4: frac100001 = 10000 - For CO: frac100004 = 2500 Since 2500 < 10000, **carbon monoxide (CO) is the limiting reagent**. Now, determine the production yield of iron based on the limiting reagent (CO): textMoles of Fe text produced = frac34 times n(CO) = frac34 times 10000 = 7500text moles Convert these moles into kilograms to find the final mass (x): textMass of Fe = frac7500 times 56text g/mol1000text g/kg = 420text kg ### Pattern Recognition Always identify the limiting reagent first by normalizing the mole quantities with their respective stoichiometric coefficients before calculating product yields. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

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