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N2(g) + 2H(g) -> N2 H4(g) What are the volumes of N2 gas and H2 gas required to form 28.5 grams of N2 H4 at 30'C and 1.50 atm?

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Answer:

Volume of N₂ = 14.76 L

Volume of H₂ = 29.52 L

Step-by-step explanation:

Given data:

Mass of N₂H₄ formed = 28.5 g

Pressure = 1.50 atm

Temperature = 30°C (30+273 = 303 k)

Volume of N₂ and H₂ needed = ?

Solution:

Chemical equation:

N₂ + 2H₂ → N₂H₄

Number of moles of N₂H₄ formed = mass/ molar mass

Number of moles of N₂H₄ formed = 28.5 g/ 32 g/mol

Number of moles of N₂H₄ formed = 0.89 mol

Now we will compare the moles of N₂H₄ with N₂ and H₂ form balance chemical equation.

N₂H₄ : N₂

1 : 1

0.89 : 0.89

N₂H₄ : H₂

1 : 2

0.89 : 2×0.89 = 1.78 mol

Volume of H₂:

PV = nRT

1.50 atm × V = 1.78 mol × 0.0821 atm.L/mol.K × 303 K

V = 44.28atm.L /1.50 atm

V = 29.52 L

Volume of N₂:

PV = nRT

1.50 atm × V = 0.89 mol × 0.0821 atm.L/mol.K × 303 K

V = 22.14 atm.L /1.50 atm

V = 14.76 L

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