Ammonia gas can be prepared by the reaction of a metal oxide such as calcium oxide with ammonium chloride. CaO(s) +2 NH, C1(s) → 2 NH3 (g) + H,O(g) + CaCl2 (s) If 102 g of CaO and 229 g of NH4 Cl are mixed, what is the limiting reactant, and what mass of NH3 can be produced? The limiting reactant is Mass of NH3

Chemistry for Engineering Students
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Chapter4: Stoichiometry
Section: Chapter Questions
Problem 4.24PAE: 4.24 Ammonia gas can be prepared by the reaction CaO(s)+2NH4Cl(s)2NH3(g)+H2O(g)+CaCl2(s) If 112 g of...
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6.
Ammonia gas can be prepared by the reaction of a metal oxide such as calcium oxide with ammonium chloride.
CaO(s) +2 NH, C1(s) → 2 NH3 (g) + H,O(g) + CaCl2 (s)
If 102 g of
CaO and 229 g of
NH4 Cl are mixed, what is the limiting reactant, and what mass of
NH3 can be produced?
The limiting reactant is
Mass of NH3
Transcribed Image Text:Ammonia gas can be prepared by the reaction of a metal oxide such as calcium oxide with ammonium chloride. CaO(s) +2 NH, C1(s) → 2 NH3 (g) + H,O(g) + CaCl2 (s) If 102 g of CaO and 229 g of NH4 Cl are mixed, what is the limiting reactant, and what mass of NH3 can be produced? The limiting reactant is Mass of NH3
Expert Solution
Step 1

The limiting reagent in a chemical reaction is a reactant that is totally consumed when the chemical reaction is completed. The amount of product formed is limited by this reagent, since the reaction cannot continue without it.

Step 2

Write the given reaction.

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

According to the stoichiometry, 1 mole of CaO reacts with 2 moles of NH4Cl.

Molar mass of CaO = 56.07 g/mol

Molar mass of NH4Cl = 53.5 g/mol

Therefore, 56.07 g of CaO reacts with 107 g (2×53.5) of NH4Cl.

Calculate the amount of NH4Cl required to react with 102 g of CaO.

Amount of NH4Cl = (107 g NH4Cl) (102 g of CaO) 56.07 g CaO                             =194.64 g of NH4Cl 

Therefore, 194.64 g of NH4Cl is required.

Amount of NH4Cl present is 229 g which is more than required, therefore, CaO is the  limiting reactant.

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