Question

A gas mixture containing 80.0 mole% N₂ and the balance n-hexane flows through a pipe at a rate of 100.0 m³/h. The pressure is 2.50 atm absolute and the temperature is 120.0°C. 752 a. what is the molar flow rate of the gas?____ kmol/h

Answer 1

The molar flow rate of the gas is 0.080 kmol/h.

Step-by-step explanation:

To find the molar flow rate of the gas, we need to calculate the number of moles of the gas flowing through the pipe in one hour. Given that the gas mixture contains 80.0 mole% N₂, we can assume that 80.0% of the gas is N₂ and the remaining 20.0% is n-hexane.

First, we need to calculate the number of moles of N₂ in the gas mixture. Since mole% is a proportion, we can calculate it as:

(80.0/100) * 100.0 m³/h = 80.0 m³/h

Next, we can convert the molar flow rate of N₂ to kmol/h by dividing by 1000:

80.0 m³/h ÷ 1000 = 0.080 kmol/h

Therefore, the molar flow rate of the gas is 0.080 kmol/h.

Answer 2

The molar flow rate of N₂ is approximately 0.1676 kmol/h meaning 0.1676 kilometers of moles of N₂ flow through the pipe every hour.

120.0°C + 273.15 K = 393.15 K.

Applying the ideal gas law:

PV = nRT

where:

P = pressure (2.50 atm)

V = volumetric flow rate (100.0 m³/h * 1 h/3600 s = 27.78 × 10
`^-^3` m³/s)

n = total moles of gas (unknown)

R = gas constant (8.3145 J/mol K)

T = temperature (393.15 K)

n_total = PV / RT

= (2.50 atm * 27.78 × 10
`^-^3` m³/s) / (8.3145 J/mol K * 393.15 K)

= 0.2095 kmol

moles of N₂ = 0.800 * n_total

=0.1676 kmol