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How many moles of H₂ are required to produce -8301 kJ of heat in the following reaction?N₂ (g) + 3 H₂ (g) → 2 NH₃ (g) ∆H° = -91.8 kJ/mol

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Step-by-step explanation:

N₂ (g) + 3 H₂ (g) ---> 2 NH₃ (g) ∆H° = - 91.8 kJ/mol

According to the ∆H° of reaction, when 1 mol of NH₃ is produced -91.8 kJ are released. We have to find the moles of H₂ nequired to produce -8301 kJ.

First we have to calculate the number of moles of NH₃ that should be produced in order to release that amount of heat.

moles of NH₃ = -8301 kJ * 1 mol/(-91.8 kJ)

moles of NH₃ = 90.42 mol

According to the coefficients of the reaction's equation 2 moles of NH₃ will be produced by 3 moles of H₂. The molar ratio between H₂ and NH₃ is 3 to 2. We can use that relationship to find the answer to our problem.

3 moles of H₂ = 2 moles of NH₃

moles of H₂ = 90.42 moles of NH₃ * 3 moles of H₂/(2 moles of NH₃)

moles of H₂ = 135.6 moles

Answer: 135.6 moles of H₂ are required.

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