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8 November, 03:51

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

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Answers (1)
  1. 8 November, 04:15
    0
    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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