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Look at this balanced chemical reaction: N2 + 3H2 2NH3

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?

Sagot :

Answer: A mass of 124457.96 g ammonia is produced by 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

Using ideal gas equation, 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 = tempertaure

Substitute 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{135000}{36.945}\\= 3654.08 mol[/tex]

According to the given equation, 1 mole of nitrogen forms 2 moles of ammonia. So, moles of ammonia formed by 3654.08 moles of nitrogen is as follows.

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

As moles is the mass of substance divided by its molar mass. So, mass of ammonia (molar mass = 17.03 g/mol) is as follows.

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

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

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