Question

what volume of H2 at NTP is required to convert 2.8g of N2 into NH3 ?​

Answer 1

N2+3H2 give rise to 2NH3

2.8÷28=volume of H2÷22.4

Volume of H2 =22.4 liters

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Answer 2

The volume of H₂ required to convert 2.8g of N₂ into NH₃ is 6.72 liters.

Given,

Mass of N₂ = 2.8 g

Condition - NTP

To Find,

The volume of H₂ required to convert 2.8g of N₂ into NH₃

Solution,

The chemical reaction for the conversion of N₂ into NH₃ is given as -

N₂ + 3H₂ → 2NH₃

According to stoichiometry -

1 Mole of N₂ requires 3 moles of H₂ to produce 2 moles of NH₃

Moles of N₂ = $\frac{Given mass }{Molar mass}$

The molar mass of N₂ = 14*2 = 28 g

Moles of N₂ = $\frac{2.8 }{28}$

Moles of N₂ = 0.1 mol

From the unitary method, we can say that -

0.1 mole of N₂ requires 0.3 moles of H₂ to produce 0.2 moles of NH₃

Moles of H₂ = 0.3 mol

At NTP, 1-mole gas has 22.4 liters of volume

Thus the volume of H₂ will be -

The Volume of H₂ = 0.3 * 22.4

The volume of H₂ = 6.72 liters

Hence, we get the Volume of H₂ as 6.72 liters.

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