In a reaction, 24.9 L of N2 reacts with excess Hy to produce NH3. How many liters of NH3

were produced? How many grams of NH3 is this? The pressure in the lab is 97.8 kPa and the

temperature was 23.7°C.

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.