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

84 g of N₂        

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