What mass of slaked lime is needed to decompose 10 g of ammonium chloride to give 100% yield? Ca(OH)2 + 2NH4Cl → CaCl2 + 2NH3 +

Question

3 years ago

10

2H2O

1 answer:

dimulka [17.4K] · Answer 1 · 2021-12-19T18:09:41+03:00

Answer is: mass of slaked lime is 6.92 grams.

Balanced chemical reaction: Ca(OH)₂ + 2NH₄Cl → CaCl₂ + 2NH₃ + 2H₂O.

m(NH₄Cl) = 10 g; mass of ammonium chloride.

M(NH₄Cl) = 14 + 1·4 + 35.5 · g/mol.

M(NH₄Cl) = 53.5 g/mol; molar mass of ammonium chloride.

n(NH₄Cl) = m(NH₄Cl) ÷M(NH₄Cl).

n(NH₄Cl) = 10 g ÷ 53.5 g/mol.

n(NH₄Cl) = 0.187 mol; amount of ammonium chloride.

From balanced chemical reaction: n(NH₄Cl) : n(Ca(OH)₂) = 2 : 1.

n(Ca(OH)₂) = 0.093 mol.

m(Ca(OH)₂) = n(Ca(OH)₂) · M(Ca(OH)₂).

m(Ca(OH)₂) = 0.093 mol · 74.1 g/mol.

m(Ca(OH)₂) = 6.92 g.