A steel cylinder contains 150.0 mol argon gas at a temperature of 25°C and a pressure of 7.04 MPa. After some argon has been used, the pressure is 2.00 MPa at a temperature of 19°C. What mass (g) of argon remains in the cylinder?

The mass of argon remaining in the cylinder is 1,740 g.

The given parameters;

The final number of moles of the gas is calculated as follows;

The molar mass of argon gas is given as;

molar mass = 40 g/mol

The mass of argon remaining in the cylinder is calculated as;

M = n x molar mass

M = 43.5 x 40

M = 1,740 g

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There remains 1737 grams of argon gas

Explanation:

Step 1: Data given

Moles of argon gas = 150.0 moles

Temperature of the argon gas = 25.0 °C = 298 K

Initial Pressure = 7.04 MPa

Final pressure = 2.00 MPa

Final temperature = 19.0 °C

Step 2: Calculate moles of argon

p*V = n*R*T

We can say:

P/(n*T) = R/V

or P1/(n1*T1) = P2/(n2*T2)

⇒ with P1 = initial pressure = 7.04 MPa = 69.4794 atm

⇒ with n1 = initial number of moles = 150.0 moles

⇒ with T1 = The initial temperature = 25.0 °C = 298 K

⇒ with P2 = the final pressure = 2.00 MPa = 19.7385 atm

⇒ with n2 = the final number of moles = TO BE DETERMINED

⇒ with T2 = the final temperature = 19.0 °C = 292 K

7.04 / (150.0 * 298) = 2.00/(n2 * 292)

1.575 *10^-4 = 2/(292n2)

292n2 = 2/1.575 *10^-4

n2 = 43.49 moles

Step 3: Calculate mass of argon

Mass = moles * molar mass

Mass = 43.49 * 39.95 g/mol

Mass argon = 1737 grams

There remains 1737 grams of argon gas

the molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

explanation:

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2