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If 276 g of dinitrogen tetroxide and 64.0 g of hydrazine are mixed, how many grams of nitrogen gas can you produce?

User Panthy
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1 Answer

9 votes

Answer:

84 g of N₂

Step-by-step explanation:

We begin from the reaction:

N₂O₄ + 2N₂H₄ → 3N₂ + 4H₂O

1st step: Find out the limiting reactant.

We convert the mass to moles.

276 g . 1mol/ 92g = 3 moles of N₂O₄

64 g . 1mol / 32g = 2 moles of hidrazine

Limiting reactant is the hydrazine. Ratio is 1:2, then for 3 moles of tetroixde I may need 6 moles of N₂H₄. (But I only have 2).

To determine the grams of produced nitrogen we see stoichiometry (2:3)

2 moles of hydrazine can produce 3 moles of N₂

Definetaly our 2 moles make 3 moles of gas.

We convert the moles to mass.

3 mol . 28g /1mol = 84 g of N₂

User Ben Pretorius
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