How many grams of NH3 can be prepared from 85.5 grams of N2 and 17.3 grams of H2?
Answer (3)
N2 + 3H2 ---> 2NH3
mass of N2 = 28g mass of H2 = 2g mass of NH3 = 17g
according to the reaction: 28g N2----------------- 3*2g H2 85,5g N2-------------------- x x = 18,32g H2 >>> so, nitrogen is excess
according to the reaction: 2 3g H2---------------------- 2 17g NH3 17,3g H2 ------------------------- x x = 98,03g NH3
answer: 98,03g of NH3
The question asks how many grams of NH3 can be prepared from specific amounts of N₂ and H₂. The calculation involves finding the limiting reactant, which is H₂, and then using stoichiometry to find the mass of NH₃ produced, which is 97.07 grams.
To determine how many grams of NH₃ can be prepared from 85.5 grams of N₂ and 17.3 grams of H₂, we need to find the limiting reactant and use stoichiometry based on the balanced chemical equation:
N₂(g) + 3H₂(g) → 2NH₃(g).
First, we need to convert the mass of N₂ and H₂ to moles using their molar masses:
1 mole of N₂ = 28.02 grams
1 mole of H₂ = 2.02 grams
Moles of N₂ = 85.5 grams / 28.02 grams/mole = 3.05 moles of N₂
Moles of H₂ = 17.3 grams / 2.02 grams/mole = 8.56 moles of H₂
From the stoichiometry of the reaction, 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃. Therefore, the limiting reactant is the one that will run out first. We can find this by comparing the mole ratio of N₂ to H₂:
For 3.05 moles of N₂, we would need 3 × 3.05 moles of H₂ = 9.15 moles of H₂. However, we only have 8.56 moles of H₂ available, which makes H₂ the limiting reactant.
According to the stoichiometry, 3 moles of H₂ produce 2 moles of NH₃. Thus:
Moles of NH₃ produced = (2/3) × 8.56 moles of H₂ = 5.71 moles of NH₃
Now we convert moles of NH₃ to grams using its molar mass (1 mole of NH₃ = 17 grams):
Mass of NH₃ produced = 5.71 moles × 17 grams/mole = 97.07 grams of NH₃
Therefore, 97.07 grams of NH₃ can be prepared from the given amounts of N₂ and H₂.
From 85.5 grams of nitrogen gas and 17.3 grams of hydrogen gas, approximately 97.07 grams of ammonia can be produced. This calculation involves determining the limiting reactant, which in this case is hydrogen. Using stoichiometry based on the balanced chemical equation, we find that hydrogen limits the amount of ammonia produced to about 5.71 moles, which translates to 97.07 grams.
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