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2N₂H₄(g) + N₂O₄(g) → 3N₂(g) + 4H₂O(g)

A reaction vessel initially contains 29.0 g N₂H₄ and 74.9 g of N₂O₄. Calculate the mass of N₂H₄ that will be in the reaction vessel once the reactants have reacted as much as possible. Assume 100% yield.

a. 15.5 g
b. 21.9 g
c. 29.0 g
d. 36.8 g

User Wolendranh
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1 Answer

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Final answer:

To find the mass of N₂H₄ that will be in the reaction vessel after the reaction, we can calculate the moles of N₂H₄ and N₂O₄, compare their mole ratio, and convert back to grams. The mass of N₂H₄ that remains is 15.5 g.

Step-by-step explanation:

To find the mass of N₂H₄ that will be in the reaction vessel after the reaction, we need to calculate the amount of N₂H₄ that reacts. We can do this by using the given masses of N₂H₄ and N₂O₄ and converting them to moles using their molar masses. Then, we compare the mole ratio of N₂H₄ to N₂O₄ from the balanced equation to determine how much N₂H₄ reacts. Finally, we convert the amount of N₂H₄ in moles back to grams using its molar mass.

The molar mass of N₂H₄ is 32.05 g/mol. The molar mass of N₂O₄ is 92.02 g/mol. Using the given masses, we have:
29.0 g N₂H₄ × (1 mol N₂H₄ / 32.05 g N₂H₄) = 0.904 mol N₂H₄
74.9 g N₂O₄ × (1 mol N₂O₄ / 92.02 g N₂O₄) = 0.814 mol N₂O₄
From the balanced equation, the mole ratio of N₂H₄ to N₂O₄ is 2:1. Therefore, we can determine that 0.814 mol N₂O₄ will react with 0.814/2 = 0.407 mol N₂H₄.
To find the mass of N₂H₄ that remains, we subtract the mass of N₂H₄ that reacts from the initial mass:
29.0 g - (0.407 mol N₂H₄ × 32.05 g/mol) = 15.5 g N₂H₄

User Jingqiang Zhang
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