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How many grams of ammonia can be produced from reacting a 450 L sample of nitrogen gas at a temperature of 450 K and a pressure of 300 atm?

How Many Grams Of Ammonia Can Be Produced From Reacting A 450 L Sample Of Nitrogen Gas At A Temperature Of 450 K And A Pressure Of 300 Atm class=

Sagot :

Answer: The mass of ammonia is 124457.96 g which can be produced from reacting a 450 L sample of nitrogen gas at a temperature of 450 K and a pressure of 300 atm.

Explanation:

Given: Volume = 450 L

Temperature = 450 K

Pressure = 300 atm

The reaction equation is as follows.

[tex]N_{2} + 3H_{2} \rightarrow 2NH_{3}[/tex]

Here, 1 mole of nitrogen reacts to give 2 moles of ammonia.

From the given data, moles of nitrogen are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into the above formula as follows.

[tex]PV = nRT\\300 atm \times 450 L = n \times 0.0821 L atm/mol K \times 450 K\\n = \frac{300 atm \times 450 L}{0.0821 L atm/mol K \times 450 K}\\= \frac{135000}{36.945}\\= 3654.08 mol[/tex]

For 3654.08 moles of nitrogen, the moles of ammonia produced is as follows.

[tex]2 \times 3654.08 mol\\= 7308.16 mol[/tex]

Therefore, mass of ammonia (molar mass = 17.03 g/mol) is calculated as follows.

[tex]Moles = \frac{mass}{molarmass}\\7308.16 mol = \frac{mass}{17.03 g/mol}\\mass = 124457.96 g[/tex]

Thus, we can conclude that the mass of ammonia is 124457.96 g which can be produced from reacting a 450 L sample of nitrogen gas at a temperature of 450 K and a pressure of 300 atm.

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