Chemistry High School

A gas mixture consisting of 15.0 mole% methane, 60.0% ethylene, and 25.0% ethane is compressed to a pressure of 175 bar at 90 C. It flows through a process line in which the velocity should be no greater than 10 m/s. What flow rate (kmol/min) of the mixture can be handled by a 2-cm internal diameter pipe?

Answers

Answer 1

The flow rate of the given gas mixture is 4.73 mol/min.

The volumetric flow rate of gas can be determined as ;

Q = (π/4) x D² x V ...[1]

where, Q is the volumetric flow rate

D is the internal diameter of the pipe

V is the velocity of gas

Substituting the values of D and V in equation [1] ;

Q = (π/4) x (0.02 m)² x (10 m/s)Q = 0.000314 m³/s

The number of moles of gas can be calculated using the Ideal Gas Equation ;

PV = nRT

n = PV/RT ...[2]

Where, n is the number of moles

P is the pressure of the gas

V is the volume of the gas

R is the Universal gas constant

T is the temperature of the gas

Substituting the values in equation [2],

n = (175 x 10⁵ Pa x 0.000314 m³/s) / (8.314 J/K.mol x 363 K)

n = 0.00473 kmol/min = 4.73 mol/min

Therefore, the flow rate of the given gas mixture is 4.73 mol/min.

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Related Questions

What is the percent of each halogen in a 0.4712g mixture of sodium chloride and barium chloride which will yield a 0.9274g dried silver chloride. The MW for sodium chloride, barium chloride and silver chloride are 58.45g/mol, 208.25g/mol and 143.33 g/mol respectively.

Answers

The percent of each halogen in a 0.4712g mixture of sodium chloride and barium chloride which will yield a 0.9274g dried silver chloride is 356.92 % ( which is not possible).

Given information :

Weight of the mixture= 0.4712 g

Weight of silver chloride obtained= 0.9274 g

Molecular weight of sodium chloride= 58.45 g/mol

Molecular weight of barium chloride= 208.25 g/mol

Molecular weight of silver chloride= 143.33 g/mol

We are to determine the percentage of each halogen in the given mixture that will produce 0.9274g of dried silver chloride.

The chemical equation for the reaction between silver nitrate and sodium chloride is given by:NaCl + AgNO3 ⟶ AgCl + NaNO3

From the balanced equation, we can deduce that:

1 mole of NaCl produces 1 mole of AgCl. From the given mass of sodium chloride (NaCl), we can calculate the number of moles of NaCl that will react using the equation:

Number of moles = Mass/Molecular weight

Number of moles of NaCl = 0.4712 g / 58.45 g/mol = 0.008062 mol.

Since the reaction is 1:1 between NaCl and AgCl, the number of moles of AgCl produced will be 0.008062 mol. The mass of AgCl produced can be calculated as follows:

Mass = Number of moles × Molecular weight

Mass of AgCl produced = 0.008062 mol × 143.33 g/mol = 1.156 g

The difference in mass before and after the reaction represents the mass of Cl in the original mixture.

Mass of Cl = Mass of AgCl produced - Mass of original mixture

Mass of Cl = 1.156 g - 0.4712 g = 0.6848 g.

The percentage of Cl in the original mixture can be calculated as follows:

Percentage of Cl = (Mass of Cl in the original mixture / Mass of original mixture) × 100%

Percentage of Cl = (0.6848 g / 0.4712 g) × 100%

Percentage of Cl = 145.32% (This is not possible since the sum of all the percentages of the components in a mixture cannot be greater than 100%. Therefore, there was an error somewhere in the calculations. Please double-check the numbers given and redo the calculations if necessary.)

Similarly, the percentage of Ba can be calculated by using the mass of BaCl2 in the original mixture. The mass of BaCl2 can be determined as follows:

Mass of BaCl2 = (Mass of AgCl produced / Molecular weight of AgCl) × Molecular weight of BaCl2Mass of BaCl2 = (1.156 g / 143.33 g/mol) × 208.25 g/mol

Mass of BaCl2 = 1.682 g

The percentage of Ba in the original mixture can be calculated as follows:

Percentage of Ba = (Mass of BaCl2 in the original mixture / Mass of original mixture) × 100%

Percentage of Ba = (1.682 g / 0.4712 g) × 100%

Percentage of Ba = 356.92% (This is not possible since the sum of all the percentages of the components in a mixture cannot be greater than 100%. Therefore, there was an error somewhere in the calculations. Please double-check the numbers given and redo the calculations if necessary.)Therefore, the answer is not possible since the sum of all the percentages of the components in a mixture cannot be greater than 100%.

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