High School

In a reaction, 24.9 L of \( \text{N}_2 \) reacts with excess \( \text{H}_2 \) to produce \( \text{NH}_3 \).

- How many liters of \( \text{NH}_3 \) were produced?
- How many grams of \( \text{NH}_3 \) is this?

The pressure in the lab is 97.8 kPa, and the temperature was 23.7°C.

Answer :

Answer:

A. 49.8L of NH3.

B. 33.83g.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is shown below:

N2 + 3H2 —> 2NH3

A. From the balanced equation above,

1L of N2 produced 2L of NH3.

Therefore, 24.9L of N2 will produce = 24.9 x 2 = 49.8L of NH3.

Therefore, 49.8L of NH3 is produced from the reaction.

B. Determination of the mass NH3 produced.

First, we shall determine the number of mole of NH3 produced. This can be obtained as follow:

Volume (V) = 49.8L

Pressure (P) = 97.8 kPa = 97.8/101.325 = 0.97atm

Temperature (T) = 23.7°C = 23.7°C + 273 = 296.7K

Gas constant (R) = 0.082atm.L/Kmol

Number of mole (n) =...?

PV = nRT

n = PV /RT

n = (0.97 x 49.8) / (0.082 x 296.3)

n = 1.99 mole

Next, we shall convert 1.99 mole to NH3 to grams. This is illustrated below:

Number of mole NH3 = 1.99 mole

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 =..?

Mass = mole x molar Mass

Mass of NH3 = 1.99 x 17

Mass of NH3 = 33.83g

Therefore, 33.83g of NH3 is produced.

Final answer:

24.9 L of N2 produces 49.8 L of NH3. Using the Ideal Gas Law, this volume of NH3 corresponds to about 2.02 moles. With a molar mass of 17 g/mol, this amounts to approximately 34.34 grams of NH3.

Explanation:

To solve this problem, we need to use the Ideal Gas Law (PV = nRT) and the balanced chemical equation for the process: N2 + 3H2 → 2NH3. According to this equation, the mole ratio between N2 and NH3 is 1:2. This means that for every 1 mole (or in this case, 24.9 L) of N2 reacted, 2 moles (or 49.8 L, because we assume standard temperature and pressure where 1 mole = 22.4 L) of NH3 are produced.

To find the mass of this NH3, we first need to find the number of moles. Using the Ideal Gas Law, we can calculate the number of moles as n = PV/RT. Given P = 97.8 kPa, V = 49.8 L, R = 8.31 (kPa L)/(K mol), and T = 23.7°C + 273.15 = 296.85 K, we find n ≈ 2.02 moles of NH3.

Finally, the molar mass of NH3 is approximately 17 g/mol, so the mass of the produced NH3 can be calculated as 2.02 moles * 17 g/mol ≈ 34.34 grams.

Learn more about Ideal Gas Law here:

https://brainly.com/question/30458409

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