what mass of slaked lime would be required to decompose completely 4g of ammonium chloride and would be the mass of each product
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question :-
what mass of slaked lime would be required to decompose completely 4g of ammonium chloride and would be the mass of each product
answer:-
What mass of slaked lime would be required to decompose completely 4 grams of ammonium chloride and what would be the mass of each product? (i) To calculate the mass of Ca(OH)2 required ot decompose 4 g NH4Cl. Thus, the mass of slaked lime required = 2.766 g
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Answer:
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 = 111
Molar 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)₂
Thus, 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.
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