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valina [46]
3 years ago
9

Hydrazine, n2h4, is used as a rocket fuel. in the reaction below, if 80.1 g of n2h4 and 92.0 g of n2o4 are allowed to react, how

many grams of excess reactant remain at the end of the reaction?
Chemistry
1 answer:
Mama L [17]3 years ago
6 0
Answer is: excess of hydrazine is 16 grams.
Chemical reaction: N₂O₄(l) + 2N₂H₄(l) → 3N₂(g) + 4H₂<span>O(g).
</span>m(N₂H₄) = 80,1 g.
m(N₂O₄) = 92,0 g.
n(N₂H₄) = m(N₂H₄) ÷ M(N₂H₄).
n(N₂H₄) = 80,1 g ÷ 32 g/mol.
n(N₂H₄) = 2,5 mol.
n(N₂O₄) = 92 g ÷ 92 g/mol.
n(N₂O₄) = 1 mol; limiting reactant.
From chemical reaction: n(N₂H₄) : n(N₂O₄) = 2 : 1.
n(N₂H₄) = 2 mol reacts.
Δn(N₂H₄) = 2,5 mol - 2 mol = 0,5 mol.
Δm(N₂H₄) = 0,5 mol · 32 g/mol = 16 g.

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The rate of increase of NH3 is 0.22M/s.

<h3>What is the balanced equation of the reaction?</h3>

The balanced equation of the reaction is given below:

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The rate of decrease of N2 is half the rate of increase of NH3.

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