The mass of NH₃ produced in the reaction of 10.5 g of nitrogen and 20.2 g
of hydrogen is 12.75 g
The balanced chemical equation is represented as follows:
3H₂ (g) + N₂ (g) → 2NH₃ (g)
Knowing the limiting reagent will assist us to ascertain the amount of Ammonia produced.
Therefore,
mass of H₂ = 20.2 g
molar mass = 2g/mol
moles = 20.2 / 2 = 10.1 moles
10.1 / 3 = 3.36666666667
mass of N₂ = 10.5 g
molar mass N₂ = 28 g /mol
moles = 10.5 / 28 = 0.375 moles
0.375 / 1 = 0.375
Therefore, nitrogen is the limiting reagent and it will determine the amount
of product formed.
28 g of N₂ gives 2(17) g of NH₃
10.5 g of N₂ will give ? of NH₃
cross multiply
mass of NH₃ produced = 10.5 × 34 / 28
mass of NH₃ produced = 357 / 28
mass of NH₃ produced = 12.75 g
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