Ammonia solvery process is used to manufacture sodium carbonate.During the following reaction 2 NH4Cl+CA( OH)2 ⬆️ CaCl2 + 2H₂O + 2NH. when 100g of ammonia chloride and 150g of calcium hydroxide are use. calculate the mass in kg of ammonia produce during the chemical reaction. calculate the excess mass in grams of one of the reactants left unreacted
Answers
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34 × 10⁻³ Kg of Ammonia
76 g of Ca(H)₂ will unreacted.
Explanation:
Given: mass of ammonium chloride = 100 g,
mass of calcium hydroxide = 150 g
number of moles NH₄Cl = 100/53.49
= 1.86moles
number of moles of Ca(OH)₂ = 150/74
= 2 moles
Reaction:
2NH₄Cl + Ca(H)₂ → CaCl₂ + 2NH₃ + 2H₂O
2 mole of NH₄Cl reacts with one mole Ca(H)₂ to give 2 moles of NH₃
But given the amount of Ca(H)₂, is 2 mole hence it is present in the excess.
Hence 1 mole = 74 g of Ca(H)₂ so 150 - 74 = 76 g of Ca(H)₂ will unreacted.
1 mole of Ca(H)₂ gives 2 moles of Ammonia
mass of NH₃ = 2 × 17 = 34 g = 34 × 10⁻³ Kg of Ammonia