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Limiting and excess reagent Watch

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

Updated: October 26, 2016
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