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tekilochka [14]
3 years ago
13

2N2H4+ N2O4———3N2+4H2O SalmaKhan99 avatar How many grams of N2 gas will be formed by reacting 100g of N2H4 and 200g of N2 Kindly

answer please
Chemistry
1 answer:
Alona [7]3 years ago
4 0

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

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Answer:

Option A

Explanation:

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3 years ago
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Molar mass:-

\\ \tt\longmapsto 14+3(1)=17g/mol

Now

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2 years ago
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Answer:

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3 years ago
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8 0
3 years ago
A chemist adds 1.80L of a 1.1/molL aluminum chloride AlCl3 solution to a reaction flask. Calculate the millimoles of aluminum ch
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Answer:

2000 millimoles of AlCl₃

Explanation:

From the question given above, the following data were obtained:

Volume of solution = 1.8 L

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Next, we shall determine the number of mole of AlCl₃ in the solution.

This can be obtained as follow:

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Thus, the chemist added 2000 millimoles of AlCl₃

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