Question

A weighed amount of sodium chloride is completely dissolved in a measured volume of 4.00 M ammonia solution at ice temperature, and carbon dioxide is bubbled in. Assume that sodium bicarbonate is formed until the limiting reagent is entirely used up. The solubility of sodium bicarbonate in water at ice temperature is 0.75 mol per liter. Also assume that all the sodium bicarbonate precipitated is collected and converted quantitatively to sodium carbonate The mass of sodium chloride in (g) is 14.88 The volume of ammonia solution in (mL) is 36.46 Calculate the following: How many grams of sodium carbonate will be produced from this precipitate??

Answer 1

son: Reaction involved (Solvay process) - Nach + cog+ NH3 + Hoo NaHCO3 + Nhy el cas - Q NaHCO Nagcoz + (Hot coop) HgCO3 * Given Nac); Weight = 14.88 gm Weight - (Moless) = 14.88 - 0.25 4 mole. : Naci 58.5(2) NH3 : Concentration = 4 mol = 4M Litre. Volume of NH3 Solution = 36.46 mL Moles of NH3 = (Molarity x Volume) (more) x (3646) Litre = 0.146 mole NaHCO3 : Solubility = 0.75 mol at ice temperature : Volume of solution - 3646 me (36.46m) Litre Nato soluble in 36-46 me = 0.75 mol 36-46 mol Litre Lite = 0.08734 moll

using Mole Concept. Naclt cog+ NH3 + H₂O NH4Cl + NaHCO3 w for t mol of Nacl. I mole of Nho is required. Nac) - 0.954 mole. He same Stoichiometric coefficient NH3 = 0.146 mole in given reaction / -NH3 is limiting reagent and - NH3 will decide the moles of product formed - Nach + co + NH3 + H₂O - NHCI . NH4Cl + + NaHCO3 cel I .P 0.146 mile Naha ): - (6.146 mol 6.146 mole] after completion of reaction Now out of 0.146 mole of NaHCO3, 0.87 34 mole of NaHCO eremains in solution and extra- motes will be ppt Moles of Natoz ppt = (0.146 – 0.2734) - = 0.11866 mole using second equation & Nattou. Na co + H2CO (Solvay process) – Mole Concept: & mole of Natico = t mole of Ngco, formed

- 0-11866 mol of Nahcoy = 0.11866 mol of Ng Cao formed Mass of No coa formed = Moles x (Polar Mass) Nacon [Mo = 106.] = (0.11860 ml) x 106 (L.) to 7/3578 gm