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At STP, how many liters of NH3 can be produced from the reaction of 6L of N2 with 6L of H2?

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Final answer:

At STP, reacting 6L of N2 with 6L of H2 following the equation N2 + 3 H2 → 2 NH3 will result in the production of 4L of NH3, as hydrogen is the limiting reactant.

Step-by-step explanation:

At STP (standard temperature and pressure), when reacting 6L of N₂ with 6L of H₂ to produce NH₃, it is important to recognize that gases at STP follows Avogadro's law, which states that equal volumes of gases at the same temperature and pressure contain an equal number of molecules.

According to the balanced chemical equation N₂ + 3 H₂ → 2 NH₃, one volume of N₂ would react with three volumes of H₂ to produce two volumes of NH₃. However, since we only have an equal volume of N₂ and H₂ (6L of each), hydrogen will be the limiting reactant. This means that after the reaction, there will be leftover nitrogen.

As per the stoichiometry of the reaction, 1L of N₂ reacts with 3L of H₂ to form 2L of NH₃.

Therefore, with 6L of H₂, we will be able to produce 4L of NH₃ before the H₂ is completely consumed.

This means that from the reaction of 6L of N₂ with 6L of H₂, we can expect to produce 4L of NH₃ at STP.

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