Stoichiometry and Redox Reactions Formulae Sheet PDF

Chem i str y | 1.23 FORMULAE SHEET 1.24 | Stoichiometr y and Redox Reactions RULES IN BRIEF The following are the definitions of ‘mole’ represented in the form of equations: Weight in g (a) Number of moles of molecules =...

Chem i str y | 1.23 FORMULAE SHEET 1.24 | Stoichiometr y and Redox Reactions RULES IN BRIEF The following are the definitions of ‘mole’ represented in the form of equations: Weight in g (a) Number of moles of molecules = Molecular weight Weight in g (b) Number of moles of atoms = Atomic weight Volume at NTP (c) Number of moles of gases = Standard molar volume (Standard molar volume is the volume occupied by 1 mole of any gas at NTP, which is equal to 22.4 litres.) No. of atoms / molecules / ions / electrons (d) Number of moles of atoms / molecules / ions / electrons = Avogadro constant (e) Number of moles of solute = Molarity × Volume of solution in litres Or No. of millimoles = Molarity × Volume in mL. Millimoles = moles 1000 (f) For a compound Mx , Ny , x moles of N = y moles of M  Chem i str y | 1.25 Solved Examples JEE Main/Boards Solving further, one gets the mass and % of the base. Milli eq. of HCl initially = 10 × 0.5 = 5 Example 1: Calculate the composition of 109% oleum. Milli eq. of NaOH consumed Sol: Let the mass of SO3 in the sample be 'w' g, then the = Milli eq.of HCl in excess = 10 × 0.2 = 2 mass of H2SO4 would be (100 – w)g. On dilution, ∴ Milli eq. of HCl consumed SO3 + H2 O → H2SO 4 80g 18g = Milli eq. of Ba(OH)2 = 5 – 2 = 3 w ∴ Eq. of Ba(OH)2 = 3/1000 = 3 × 10−3 Moles of SO3 in oleum = = Moles of H2SO4 formed after dilution. 80 Mass of Ba(OH)2 = 3 × 10−3 (171/2) = 0.2565 g 98w % Ba(OH)2 = (0.2565/20) × 100 = 1.28% ∴ Mass of H2SO4 formed on dilution = 80 Total mass of H2SO4 present in oleum after dilution Example 3: One litre of mixture of CO and CO2 is passed 98w through red hot charcoal in tube. The new volume = + (100 − w)= 109 ; w = 40 becomes 1.4 litre. Find out % composition of original 80 mixture by volume. All measurements are made at same Thus oleum sample contains 40% SO3 and 60% H2SO4. P and T. Example 2: 20g of a sample of Ba(OH)2 is dissolved in Sol: Assuming the mixture contents as a and b, the 10 mL. of 0.5 N HCl sol. The excess of HCl was titrated reaction is framed and values are laid down. with 0.2 N NaOH. The volume of NaOH used was 10 cc. Let the mixture contains Calculate the percentage of Ba(OH)2 in the sample. CO = a litre; CO2 = b litre Sol: The titration principle is applied wherein milli- ∴ a + b = 1 …(i) equivalents of the neutralization reactions is calculated.