Ammonia gas can be prepared by the reaction of metal oxide such as calcium oxide with ammonium chloride: cao (s) 2nh4cl 2nh3 (g) h2o(l) cacl2 (s) when 112 g of cao and 224 g of nh4cl react, 50 g of nh3 is produced. calculate the percentage yield for the reaction. show all the steps in the calculation.

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26-08-2022
Chemistry

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Ammonia gas can be prepared by the reaction of metal oxide such as calcium oxide with ammonium chloride: cao (s) 2nh4cl 2nh3 (g) h2o(l) cacl2 (s) when 112 g of cao and 224 g of nh4cl react, 50 g of nh3 is produced. calculate the percentage yield for the reaction. show all the steps in the calculation.

Answer :

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nandinipareek nandinipareek · Answer 1 · 2022-09-03T10:37:58-04:00

The mass of NH3 reaction yield, calculated as a percentage, is : 68.02 grams.

Given data :

mass of CaO = 112 g

mass of NH4Cl = 224 g

molar mass of CaO = 56.08 g/mol

molar mass of NH4Cl = 53.49 g/mol

molar mass of NH3 = 17.03 g/mol

1.

Balanced chemical reaction is given by :

CaO(s) + 2NH4Cl(s) = 2NH3(g) + H2O(g) + CaCl2(s)

2.

Calculation for moles of CaO :

moles = mass / molar mass

moles of Cao = 112g / 56.08g/mol

moles of CaO = 1.997moles

3.

Calculation for moles of NH4Cl :

moles = mass / molar mass

moles of NH4Cl = 224g / 53.49g/mol

moles of NH4Cl = 4.188 moles

4.

Identifying limiting reagent:

Reacting moles of Cao = 1.997 mol / 1 = 1.997 mol

Reacting moles of NH4Cl = 4.188 mol / 2 = 2.09 mol

So, moles of CaO are comparitively lesser than moles of NH4Cl.

Therefore, CaO is the limiting reagent.

5.
Calculation for the mass of NH3:

1 moles of Cao produces 2 moles of NH.3.

1.997 moles of CaO produces 3.994 moles of NH3.

mass = moles x molar mass

mass of NH3 = 3.997 mol * 17.03 g/mol

mass of NH3 = 68.02 g

= 68.0 g Approx.

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