Answer:
131.26 g
Explanation:
From the balanced equation,
2 moles of N₂H₄reacts with 1 mole of N₂O₄ to give 3 moles of N₂
Now number of moles of N₂H₄ present in 100 g N₂H₄ is n = 100 g/molar mass N₂H₄.
Molar mass N₂H₄ = 2 × 14.01 g/mol + 1 × 4 g/mol = 28.02 g/mol + 4 g/mol = 32.02 g/mol
n₁ = 100/32.02 = 3.123 mol
Also
Now number of moles of N₂O₄ present in 200 g N₂O₄ is n = 200 g/molar mass N₂O₄.
Molar mass N₂O₄ = 2 × 14.01 g/mol + 16 × 4 g/mol = 28.02 g/mol + 64 g/mol = 92.02 g/mol
n₂ = 200/92.02 = 2.173 mol
Since the mole ratio of N₂H₄ to N₂O₄ is 2 : 1, We require 2 × 2.173 mol N₂H₄ to react with 2.173 mole N₂O₄
Number of moles of N₂H₄ required is 4.346. But the number of moles of N₂H₄ present is 3.123 so N₂H₄ is the limiting reagent.
So, from the equation, 2 moles of N₂H₄ produces 3 moles of N₂
Therefore number of mole N₂ = 3/2 moles of N₂H₄ = 3/2 × 3.123 mol = 4.6845 mol
From n = m/M where n = number of moles of nitrogen gas = 4.6845 mol and M = molar mass of nitrogen gas = 28.02 g/mol and m = mass of nitogen gas.
m = nM = 4.6845 mol × 28.02 g/mol = 131.26 g
So the mas of nitrogen gas produced is 131.26 g
