Question

N2+3H2—> 2NH3 What volume of NH3 at STP is produced if 25.0g of N2 is reacted with an excess of H2 ?

Answer 1

The volume of NH3 produced at STP from 25.0 g of N2, when reacted with an excess of H2, is 40.0402 liters.

Step-by-step explanation:

To calculate the volume of NH3 produced at STP (Standard Temperature and Pressure) from 25.0 g of N2 reacting with an excess of H2, first, we need to determine the moles of N2:

Molar mass of N2 = 28.0 g/mol

Moles of N2 = mass / molar mass = 25.0 g / 28.0 g/mol = 0.8929 mol

Using the balanced chemical equation:

N2(g) + 3H2(g) → 2NH3(g)

We see that 1 mole of N2 produces 2 moles of NH3. Therefore, 0.8929 moles of N2 will produce 2 * 0.8929 moles of NH3, which is 1.7858 moles.

At STP, 1 mole of gas occupies 22.4 liters. The volume of NH3 produced will therefore be:

Volume of NH3 = moles of NH3 * volume of 1 mole at STP

Volume of NH3 = 1.7858 mol * 22.4 L/mol = 40.0402 L

Answer 2

The answer to your question is 40 L of NH₃

Step-by-step explanation:

Data

Volume of NH₃ = x

mass of N₂ = 25 g

mass of H₂ = excess

Balanced chemical reaction

N₂ + 3H₂ ⇒ 2NH₃

Process

1.- Find the molar mass of N₂ and NH₃

N₂ = 14 x 2 = 28g

2NH₃ = 2[ 14 + 3] = 34 g

2.- Write a proportion to solve this problem

28 g of N₂ --------------- 34 g of NH₃

25 g of N₂ ------------- x

x = (25 x 34)/28

x = 30.36 g of NH₃

3.- Calculate the volume of NH₃

17 g of NH₃ -------------- 22.4 L

30.36 g of NH₃ -------- x

x = (30.36 x 22.4) / 17

x = 40 L