What would be the weight of the slaked lime required to decompose 8.0 g of ammonium chloride?
Answers
Answered by
49
Here we have to calculate the weight of the slaked lime required to decompose 8.0 g of ammonium chloride.
Solution:
The chemical equation for this reaction is:
2NH₄Cl + Ca(OH)₂ → 2NH₃ + 2H₂O + CaCl₂
Mass of 2(NH₄Cl) = 2 (14+4+35.5) = 107 g
Mass of Ca(OH)₂ = 40 + 2 (1 + 16) = 74 g
Above reaction shows that,
107 g of NH₄Cl is decomposed by 74 g of Ca(OH)₂
So, 1 g of NH₄Cl is decomposed by 74/107 = 0.692 g of Ca(OH)₂
Now, 8 g of NH₄Cl is decomposed by 0.692 x 8 = 5.536 g of Ca(OH)₂
This is the required answer.
Hopefully it is helpful to you. Thanks.
Solution:
The chemical equation for this reaction is:
2NH₄Cl + Ca(OH)₂ → 2NH₃ + 2H₂O + CaCl₂
Mass of 2(NH₄Cl) = 2 (14+4+35.5) = 107 g
Mass of Ca(OH)₂ = 40 + 2 (1 + 16) = 74 g
Above reaction shows that,
107 g of NH₄Cl is decomposed by 74 g of Ca(OH)₂
So, 1 g of NH₄Cl is decomposed by 74/107 = 0.692 g of Ca(OH)₂
Now, 8 g of NH₄Cl is decomposed by 0.692 x 8 = 5.536 g of Ca(OH)₂
This is the required answer.
Hopefully it is helpful to you. Thanks.
Answered by
2
Answer:
Hi, I support Real Madrid, you should also support them Please.
Attachments:
Similar questions