Final answer:
To produce 18.5 grams of ammonia with an 86% yield, 2.812 grams of hydrogen gas are required, taking into account the stoichiometry of the reaction and the molar masses of the compounds involved.
Step-by-step explanation:
To calculate the amount of hydrogen gas (H₂) required to produce 18.5 grams of ammonia (NH₃) with an 86% yield, we must first convert the mass of NH₃ into moles using its molar mass.
The molar mass of NH₃ is 17 g/mol, so 18.5 grams is equivalent to 18.5 g / 17 g/mol = 1.09 moles of NH₃. According to the chemical equation N₂(g) + 3 H₂(g) → 2 NH₃(g), it takes 3 moles of H₂ to produce 2 moles of NH₃.
To find the moles of H₂ required for the theoretical yield of 1.09 moles of NH₃, we set up a proportion: (3 moles H₂ / 2 moles NH₃) × 1.09 moles NH₃ = 1.635 moles H₂.
Considering an 86% yield, we multiply the theoretical moles of H₂ by 0.86: 1.635 moles × 0.86 = 1.406 moles H₂. Finally, to convert moles of H₂ to grams, we use the molar mass of H₂, which is approximately 2 g/mol. Therefore, 1.406 moles × 2 g/mol = 2.812 grams of H₂.