f all Your question is wrong......it should be ...What mass of SLAKED lime would be required to decompose COMPLETELY 4 g of NH4Cl and what would be the mass of each product?now the answer is ....The reaction for the DECOMPOSITION of the Slaked lime and Calcium Hydroxide is ⇒2NH₄Cl + Ca(OH)₂ --------→ CaCl₂ + 2H₂O + 2NH₃↑Molar mass of the NH₄Cl = 53.5 g/mole.Molar mass of the Ca(OH)₂ = 74 g/mole.Molar mass of CaCl₂ = 111Molar mass of water = 18 g/mole.Molar mass of AMMONIA = 17 g/mole.From the Reaction,∵ 2 × 53.5 grams of NH₄Cl requires 74 g of the Ca(OH)₂∴ 1 g of NH₄Cl requires 74/107 g of the Ca(OH)₂.∴ 4 g of the NH₄Cl requires 0.692 × 4 g of the Ca(OH)₂.    = 2.768 g of the Ca(OH)₂Thus, Mass of the Calcium Hydroxide is 2.768 grams.For the Mass of the CaCl₂.∵ 2 × 53.5 g of the AMMONIUM Chloride produces 111 g of Calcium Chloride.∴ 1 g of the ammonium ch. produces 111/107 g of CaCl₂.∴ 4 g of NH₄Cl produces 1.04 × 4 g = 4.16 g of CaCl₂ For the mass of water,107 g of Ammonium chloride produces 2 × 18 g of Water.∴ 4 g of the Ammonium chloride produces 36/107 × 4 = 1.356 g of water.For the Mass of Ammonia,107 g of Ammonium Chloride produces 2 × 17 g of Ammonia.∴ 1 g of Ammonium Chloride produces 34/107 g of Ammonia.∴ 4 g of NH₄Cl produces 34/107 × 4 = 1.27 g of NH₃.Hope it helps.