How many grams of NH3 can be produced theoretically from the reaction of 4.00g of N2 and 4.00g of H2?

Answer 1

The balanced chemical reaction is:

N₂ (g) + 3 H₂ (g) → 2 NH₃ (g)

Use stoichiometry to convert 4.00 g of N₂ and H₂ to moles of NH₃ to first determine the limiting reagent. Molar mass of N₂ = 28.00 g/mole (NOTE! The other tutor incorrectly used 14 g/mol in his calculations); molar mass of H₂ = 2.02 g/mole; mole ratio of N₂ to NH₃ is 1:2; mole ratio of H₂ to NH₃ is 3:2.

4.00 g N₂ x (1 mol N₂ / 28.00 g N₂) x (2 mol NH₃ / 1 mol N₂) = 0.286 mol NH₃

4.00 g H₂ x (1 mol H₂ / 2.02 g H₂) x (2 mol NH₃ / 3 mol H₂) = 1.32 mol NH₃

Since N₂ produces less NH₃, it is the limiting reagent. Now convert 0.286 mol NH₃ to grams using its molar mass (17.03 g/mol).

0.286 mol NH₃ x (17.03 g NH₃ / 1 mol NH₃) = 4.87 g NH₃

Answer 2

balance the chemical reaction

N₂ + 3H₂------ 2 NH₃

convert the masses into grams

4.00g N₂=4/14=0.286 mole

4.00g H₂=4/2= 2 moles

find the limiting reactant

0.286 mole of N₂ give 0.572 mole of NH₃

2 moles of H₂ give 4/3 moles of NH₃

the limiting reactant is N₂

the NH₃ produced is 0.572 moles

convert the moles into grames so 0.572 moles NH₃ x17g/1mole = 9.724 g

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