Question

calculate the number of gram of ammonia gas (NH3) contained in the a 3.0L vessel at 305 K with a pressure of 1.50 atm

Answer 1

Answer: The number of gram of ammonia gas (
`NH_(3)`) contained in the a 3.0L vessel at 305 K with a pressure of 1.50 atm is 3.048 g.

Step-by-step explanation:

Given: Volume = 3.0 L

Temperature = 305 K

Pressure = 1.50 atm

Formula used to calculate the number of moles is 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 above formula as follows.

`PV = nRT\\1.50 atm * 3.0 L = n * 0.0821 L atm/mol K * 305 K\\n = (1.50 atm * 3.0 L)/(0.0821 L atm/mol K * 305 K)\\= (4.5)/(25.0405)\\= 0.179 mol`

It is known that moles is the mass of substance divided by its molar mass.

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

`Moles = (mass)/(molar mass)\\0.179 mol = (mass)/(17.03 g/mol)\\mass = 3.048 g`

Thus, we can conclude that the number of gram of ammonia gas (
`NH_(3)`) contained in the a 3.0L vessel at 305 K with a pressure of 1.50 atm is 3.048 g.