N2(g) + 3H2(g) ------> 2NH3(g)
Stoichiometrically, 1mole of N2 react with 3moles of H2
Number of moles = mass / molar mass
number of moles of N2 = 1200g/28.014g/mol = 42.836mol
number of moles of H2 = 590g/2.016g/mol = 292.66mol
292.66moles of H2 require 97.553moles of N2 but available moles of N2 is 42.836. So,
Limiting reagent is H2
Stoichiometrically, 3moles of H2 gives 2moles of NH3
moles of NH3 obtained by 42.836moles of H2 = (2/3)×42.836mol = 28.557mol
Mass of NH3 that can be maximum produced = 28.557mol × 17.031g/mol
= 486.34 g
Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:...
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