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Consider the following reaction: 3PbCl_2 + 2(NH_4)_3PO_4 rightarrow Pb_3(PO_4)_2 + 6NH_4Cl If 72 mathrm~mmol of PbCl_2 is mixed with 50 mathrm~mmol of (NH_4)_3PO_4, then amount of Pb_3(PO_4)_2 formed is ... mmol. (nearest integer)

Numerical Answer Type:
Enter a numerical value Answer: 24 to 24 +4 marks

Solution & Explanation

### Related Formula textMoles of Product = textMoles of Limiting Reagent times fractextStoichiometry of ProducttextStoichiometry of Limiting Reagent ### Core Logic From the balanced chemical equation: 3 text moles of PbCl_2 text react with 2 text moles of (NH_4)_3PO_4. Let's find the limiting reagent (L.R.) by dividing given millimoles by stoichiometric coefficients: For PbCl_2: frac723 = 24 For (NH_4)_3PO_4: frac502 = 25 Since 24 < 25, PbCl_2 is the limiting reagent and will completely consume. ### Step 1: Calculate Product Moles Moles of Pb_3(PO_4)_2 formed depends entirely on PbCl_2. 3 mmol of PbCl_2 produces 1 mmol of Pb_3(PO_4)_2. Therefore, 72 mmol of PbCl_2 will produce: frac13 times 72 = 24 mathrm~mmol of Pb_3(PO_4)_2. ### Pattern Recognition Always identify the Limiting Reagent by taking the ratio n / textcoefficient. The smallest ratio dictates the extent of the reaction. ### 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 5

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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