Final answer:
To produce 119 grams of ammonia, 21 grams of hydrogen are needed when reacted with excess nitrogen according to the balanced chemical equation.
Step-by-step explanation:
The student is asking how many grams of hydrogen (H₂) are necessary to produce 119 grams of ammonia (NH₃) given that nitrogen (N₂) is in excess. The balanced chemical equation for this reaction is: N₂(g) + 3H₂(g) → 2NH₃(g).
First, we need to calculate the moles of ammonia produced using its molar mass (17 g/mol). With 119 g of ammonia, that is 119 g / 17 g/mol = 7 moles of NH₃.
From the balanced equation, we have a mole ratio of 2 moles of NH₃ for every 3 moles of H₂. Therefore, to find the moles of H₂ needed, we set up a proportion: (7 mol NH₃) x (3 mol H₂) / (2 mol NH₃) = 10.5 moles of H₂. Lastly, using the molar mass of hydrogen (2 g/mol), we multiply by the moles of hydrogen to get the mass: 10.5 moles H₂ x 2 g/mol = 21 grams of hydrogen are needed.