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)

Solution & Explanation

### 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

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Some Basic Concepts Of Chemistry

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Q83 jee_main_2024_27_jan_morning Stoichiometry
Mass of methane required to produce 22text g of textCO_2 after complete combustion is textquadquad g. (Given Molar mass in textg mol^-1: textC=12.0, textH=1.0, textO=16.0)
Numerical Answer. Answer: 8 to 8

Solution

### Related Formula Balanced combustion chemical equation: textCH_4 + 2textO_2 rightarrow textCO_2 + 2textH_2textO textMoles = fractextMasstextMolar Mass ### Step 1: Determine moles of product generated textMolar Mass of CO_2 = 12 + (2 times 16) = 44text g mol^-1 textMoles of CO_2 text produced = frac2244 = 0.5text moles ### Step 2: Relate to input mass via stoichiometry metrics From the balanced equation, 1text mole of CH_4 produces 1text mole of CO_2. textRequired Moles of CH_4 = 0.5text moles textMolar Mass of CH_4 = 12 + (4 times 1) = 16text g mol^-1 textMass of CH_4 = 0.5 times 16 = 8text g ### Pattern Recognition 22text g of CO_2 is exactly half a mole. By stoichiometry ratios, half a mole of methane is needed, which translates to 8text g. ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q88 jee_main_2024_30_jan_morning Mole Concept
0.05text cm thick coating of silver is deposited on a plate of 0.05text m^2 area. The number of silver atoms deposited on plate are ________ times 10^23. (At mass Ag=108, d=7.9text g cm^-3)
Numerical Answer. Answer: 11 to 11

Solution

### Related Formula textVolume = textArea times textThickness textMass = textDensity times textVolume textMoles = fractextMasstextMolar Mass textNumber of Atoms = textMoles times N_A ### Step 1: Calculate Volume of Coating Area = 0.05 text m^2 = 0.05 times 10^4 text cm^2 = 500 text cm^2 Thickness = 0.05 text cm Volume = 500 text cm^2 times 0.05 text cm = 25 text cm^3 ### Step 2: Calculate Mass and Moles Mass = Volume times Density = 25 text cm^3 times 7.9 text g/cm^3 = 197.5 text g Moles of Ag = frac197.5108 = 1.8287 text moles ### Step 3: Calculate Number of Atoms textNumber of Atoms = 1.8287 times 6.022 times 10^23 = 11.01 times 10^23 Rounding to nearest integer, we get 11. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry Class 12 Chemistry: Electrochemistry
Q62 jee_main_2024_31_jan_evening Stoichiometry and Calculations
A sample of CaCO_3 and MgCO_3 weighed 2.21text g is ignited to constant weight of 1.152text g. The composition of mixture is: (Given molar mass in mathrmg\,mol^-1 mathrmCaCO_3:100, MgCO_3:84)
  • A. 1.187mathrm~gmathrm~CaCO_3 + 1.023mathrm~gmathrm~MgCO_3
  • B. 1.023mathrm~gmathrm~CaCO_3 + 1.023mathrm~gmathrm~MgCO_3
  • C. 1.187mathrm~gmathrm~CaCO_3 + 1.187mathrm~gmathrm~MgCO_3
  • D. 1.023mathrm~gmathrm~CaCO_3 + 1.187mathrm~gmathrm~MgCO_3

Solution

### Related Formula mathrmCaCO_3(s) xrightarrowDelta mathrmCaO(s) + mathrmCO_2(g) mathrmMgCO_3(s) xrightarrowDelta mathrmMgO(s) + mathrmCO_2(g) ### Core Logic Let the weight of mathrmCaCO_3 be xtext g. Then, the weight of mathrmMgCO_3 = (2.21 - x)text g. Moles of mathrmCaCO_3 decomposed = Moles of mathrmCaO formed. fracx100 = textMoles of CaO formed textWeight of CaO formed = fracx100 times 56 Moles of mathrmMgCO_3 decomposed = Moles of mathrmMgO formed. frac(2.21 - x)84 = textMoles of MgO formed textWeight of MgO formed = frac2.21 - x84 times 40 ### Step 1: Setting up the Equation The total weight of the residue (mathrmCaO + mathrmMgO) is given as 1.152text g. frac2.21 - x84 times 40 + fracx100 times 56 = 1.152 ### Step 2: Solving for x frac88.4 - 40x84 + 0.56x = 1.152 1.0523 - 0.4761x + 0.56x = 1.152 0.0839x = 0.0997 x = 1.188text g So, weight of mathrmCaCO_3 approx 1.187text g (accounting for rounding) Weight of mathrmMgCO_3 = 2.21 - 1.188 = 1.022text g approx 1.023text g. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q83 jee_main_2024_31_jan_morning Stoichiometry
Number of moles of methane required to produce 22g CO_2(g) after combustion is x times 10^-2 moles. The value of x is
Numerical Answer. Answer: 50 to 50

Solution

### Step 1: Stoichiometric Equation CH_4(g) + 2O_2(g) rightarrow CO_2(g) + 2H_2O_(l) 1 mole of CH_4 produces 1 mole of CO_2. ### Step 2: Moles Calculation Molar mass of CO_2 = 12 + 2(16) = 44 text g/mol n_CO_2 = fractextMasstextMolar mass = frac2244 = 0.5 text moles Since 1 mole of CH_4 produces 1 mole of CO_2, the moles of CH_4 required is 0.5 moles. ### Step 3: Finding x 0.5 text moles = 50 times 10^-2 text moles x = 50 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

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