What mass of slaked lime is needed to decompose 10 g of ammonium chloride to give 100% yield? Ca (OH) 2 + 2NH4Cl → CaCl2 + 2NH3 + 2H2O

Question

Chemistry

Aggie

16 August, 12:25

What mass of slaked lime is needed to decompose 10 g of ammonium chloride to give 100% yield? Ca (OH) 2 + 2NH4Cl → CaCl2 + 2NH3 + 2H2O

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Dana Gillespie · Answer 1

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.