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# Limiting and excess reagent

1. What is the maximum mass of ammonia that can be recovered from 2.00 tonnes(2000000g) of ammonium chloride and 0.500 tonnes of calcium oxide?
balanced eq- 2NH4Cl + CaO --> CaCl2 + H20+ 2NH3
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Just quoting in Danny Dorito so she can move the thread if needed
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(Original post by Danny Dorito)
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3. 2NH4Cl + CaO --> CaCl2 + H20+ 2NH3

First work out mass no of NH4Cl: N = 14
H4 = 4
Cl = 35.5
________
TOTAL = 53.5
2NH4Cl = 107

Mass no of CaO Ca = 40
O = 16
_______
TOTAL = 56

So 107 g of NH4CL would react with 56g of CaO i.e. tehere is a bit less than double mass of NH4Cl compared to CaO

We have 2 tonnes of NH4CL and 0.5 tonnes of CaO; here NH4CL is 4 times as much as CaO, so NH4Cl is in excess. We use the mass of CaO = 0.5 tonnes to calculate amount of NH3 gas recoverable, because CaO is the limiting factor.

Mass No of NH3 = 14 + 3 = 17But 2 moles are produced = 34 g

56 g CaO gives 34 g NH3

So 0.5 tonnes of Cao gives 500 kg/56 X 34 = 303.6 kg of ammonia

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