Ammonia, NH3, is produced by the reaction N2 (g) + 3H2 (g) → 2NH3 (g). if 3.50 mol of nitrogen is present, what volume of NH3 at 25.0°C and 1.15atm of pressure can be made?

a

0.312 L

b

74.5 L

c

157 L

d

149 L

Option D. 149L

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

N2 (g) + 3H2 (g) → 2NH3 (g)

Step 2:

Determination of the number of mole of NH3 produced by the reaction of 3.50 mole of N2. This is illustrated below:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 3.5 moles of N2 will react to produce = 3.5 x 2 = 7 moles of NH3.

Therefore, 7moles NH3 is produced.

Step 3:

Determination of the volume of NH3 produced from the reaction. This can be achieved as shown below:

Data obtained from the question include:

Temperature (T) = 25.0°C = 25.0°C + 273 = 298K

Pressure (P) = 1.15 atm

Number of mole (n) = 7 moles

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

Volume (V) =..?

Applying the ideal gas equation to obtain the volume:

PV = nRT

Divide both side by P

V = nRT /P

V = (7 x 0.082 x 298)/1.15

V = 149 L

Therefore, the volume of NH3 produced from the reaction is 149L

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