Question

15. How many grams of silver chloride could be formed when 200 mL of 0.50 M AgNO, solution is reacted with an excess of NaCl. AgNO3(aq) + NaCl (aq) → AgCl (s) + NaNO3(aq) Page 4

Answer 1

Reaction, Ag Nozcaq) + NaCl (aq) The equation is balanced. → Agai(s) + NaNO3(aq) رقيقN We are given 200 ml of 0.5M> AgNO3 Molecular weight of Ag No3 = 169.8731 gråtam :: 1 mole of Agnozcaq) reacts with esccess of Nace. So, the first one to get exchausted is AgNO3. Therefore, it is the limiting agent i.e. AgNO3. Now, from the formula, Molarity = moles of solute given man x_1000 - volume of solvent (in e molecularmass volume of solvent mu) :. reaction 0.5 = given mars x 1000 5 169.8731 200 given mass or required mass for the = 0.5X 169.8731 grams 5 = 84.93655 grams = 16.98731 grams. Now, as per stoichiometry 1 mole of AgNO₃ reacts with 1 mole of Nace to give 1 mole of Aga and 1 mole of NaNO₃ . 16.98731 = Ool moles of Ag NO₂ will give Ool mole of Agre 169.8731 molar les man :: grams of Agce = moleox" molar mass 6 weight anmoliculas mars = 0.1 roles x 143.32g Smolar mass of tooo = 143.32 glmal moles - Aga = 14-332grams 14.332 geams is the required mass of Age.

Reaction, MgSO4 (aq) + 2HNO3(aq) → Mg (NO3)2 (aq) + H2SO4 (aq) The reaction is balanced We are given 0 0.15M of Mg Son Molar mass of MgSO4 = 120.37 glmal 0 0.2L ,0.3M of HNO₂ : Molar mass of HNO₃ = 63.01 g/mol from the boumula, Molarity = mass 1000 molecular mano volume of solvent (me) :. Fou HNO3 0.3 = mass {0-22= 200me} - 63.01 200 mass = 0.3 x63.01 = 3. 78 gram 5 XLDOO :: moles e HNO3 = mass = 3.78 g = 0.0599 mol molar mano 63. olg/mol Now, as per stoichiometry reacts with a moles of HNO₃ to give I mole of Masou 1 mole of Mg(NO3)2 and 1 mole of H₂SO4. : 0.0599 mol of HNO₂ will react with MgSO4. Now, Molarity of Mg so 4 = 0.15M from formula ① , 0.15 = X 0.0599 x 1000 volume of solvent (me) .. volume of My Soy Solution = {x0.0599 x 1000 .= 29.95 me 0.15 = 199.66 ml .: Volume in Litre = 0.19966 L MgSoy Solution 0.15
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