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Consider the unbalanced reaction,
H₂ + N₂ ===> NH₃
Balance it to get
3 H₂ + N₂ ===> 2 NH₃
This tells you that for every 3 moles of H₂ and 1 mole of N₂ among the reactants, 2 moles of NH₃ are produced.
Look up the molar masses of each component element:
• H : 1.008 g/mol ===> H₂ : 2.016 g/mol
• N : 14.007 g/mol ===> N₂ : 28.014 g/mol
• NH₃ : 17.031 g/mol
Convert the given mass to moles for each reactant:
• H₂ :
(5.22 kg) × (1000 g/kg) × (1/2.016 mol/g) ≈ 2589.29 mol
• N₂ :
(31.5 kg) × (1000 g/kg) × (1/28.014 mol/g) ≈ 1124.44 mol
Now,
(2589.29 mol H₂) : (1124.44 mol N₂) ≈ (2.30 mol H₂) : (1 mol N₂)
so that the amount of N₂ consumed is
(2589.29 mol H₂) × (1 mol N₂) / (3 mol H₂) ≈ 863.095 mol N₂
leaving an excess of about 261.343 mol N₂, and producing
(2589.29 mol H₂) × (2 mol NH₃) / (3 mol H₂) ≈ 1726.19 mol NH₃
Convert this to a mass :
(1726.19 mol) × (17.031 g/mol) × (1/1000 kg/g) ≈ 29.3987 kg
So, up to 29.4 kg of NH₃ can be synthesized.