A gas containing methane, ethane, and carbon dioxide is...

A gas containing methane, ethane, and carbon dioxide is analyzed with a gas chromatograph (GC) and flame ionization detector (FID): the GC separates the components of the gas, and the FID registers signals proportional to the amount of each hydrocarbon (but not $\mathrm{CO}_{2}$ ) in its sample chamber. The output of the FID is as follows: The area under each peak is proportional to the number of carbon atoms in the sample, so that 1 mol of ethane would yield a peak with twice the area of a peak corresponding to 1 mol of methane. This fuel is being burned with air in a continuous combustion chamber. The molar feed ratio of air to fuel was supposed to be $7: 1,$ but you suspect the air flowmeter is not functioning properly. To check it, you take a 0.50-mol sample of the product gas and pass it through a condenser, which condenses essentially all of the water in the sample. The condensate (which can be assumed to be pure water) is weighed and found to have a mass of 1.134 g. The dry gas leaving the condenser is analyzed and found to contain no hydrocarbons, no CO, and 11.9\% CO $_{2}$. (a) Calculate the molar composition (component mole fractions) of the fuel gas and the desired percent excess air. (b) Calculate the actual molar feed ratio of air to fuel and the actual percent excess air.

A gas containing methane, ethane, and carbon dioxide is analyzed with a gas chromatograph (GC) and flame ionization detector (FID): the GC separates the components of the gas, and the FID registers signals proportional to the amount of each hydrocarbon (but not $\mathrm{CO}_{2}$ ) in its sample chamber. The output of the FID is as follows:
The area under each peak is proportional to the number of carbon atoms in the sample, so that 1 mol of ethane would yield a peak with twice the area of a peak corresponding to 1 mol of methane. This fuel is being burned with air in a continuous combustion chamber. The molar feed ratio of air to fuel was supposed to be $7: 1,$ but you suspect the air flowmeter is not functioning properly. To check it, you take a 0.50-mol sample of the product gas and pass it through a condenser, which condenses essentially all of the water in the sample. The condensate (which can be assumed to be pure water) is weighed and found to have a mass of 1.134 g. The dry gas leaving the condenser is analyzed and found to contain no hydrocarbons, no CO, and 11.9\% CO $_{2}$.
(a) Calculate the molar composition (component mole fractions) of the fuel gas and the desired percent excess air.
(b) Calculate the actual molar feed ratio of air to fuel and the actual percent excess air.

Key Concepts

Condensation and Water Analysis in Combustion Products

The analysis of water produced during combustion is often performed by condensing the water vapor from the combustion gases. Measuring the condensed water provides crucial information about the amount of hydrogen in the fuel and serves as an indirect method to confirm the combustion completeness and the stoichiometry of the reaction.

Combustion Stoichiometry

Combustion stoichiometry involves the quantitative analysis of reactants and products in a combustion process. By applying the law of conservation of mass and balancing chemical equations, one can determine the requisite air-to-fuel ratios, the amounts of combustion products formed, and evaluate the completeness of the combustion reaction.

Excess Air in Combustion

Excess air refers to the additional amount of air supplied to a combustion process beyond the stoichiometric requirement. It is important for ensuring complete combustion, yet too much excess air can lead to energy losses. Calculating the percent excess air helps in optimizing the efficiency of the combustion system and ensuring that emissions are controlled.

Gas Chromatography

Gas chromatography is an analytical technique used to separate the components of a mixture based on their differing interactions with a stationary phase inside a column. This separation allows for the qualitative and quantitative analysis of each individual component, providing essential information about the composition of complex gas mixtures.

Flame Ionization Detection (FID)

The flame ionization detector is a common instrument coupled with gas chromatography that detects organic compounds by burning them in a flame and measuring the resulting ions. Its response is generally proportional to the number of carbon atoms present, making it especially useful for quantifying hydrocarbons in a sample.

Transcript

00:02 For this question, we have methane reacting with oxygen provided by air.

00:09 There are several assumptions that we need to make in order to appropriately answer this question with the information that's provided.

00:15 We need to assume that at constant temperature, that the number of moles is proportional to the pressure times the volume.

00:25 If the pressure increases or the volume increases, we have an increase in the number of moles.

00:33 Moles at this constant temperature.

00:37 We don't know the temperature, so we need to make this assumption.

00:41 We also have a chemical reaction of methane reacting with oxygen, producing carbon dioxide and water.

00:49 To balance this chemical reaction, we need to put a two here, so we have four hydrogens on both sides, and then a two here, so we have four oxygens on both sides.

00:59 This will become important as we consider how much air is required to react with the methane.

01:08 If we are providing methane at 200 liters per minute at 1 .50 atmospheres, then the pressure volume will be 200 liters per minute, multiplied by the pressure of 1 .50 atmospheres.

01:33 This gives us 300 liter atmospheres per minute.

01:41 So if we are providing 300 liter atmospheres per minute, this now being proportional to the number of moles, we need to recognize that we need twice that based on the stoichiometry of oxygen.

02:01 But in addition, the problem says we're going to use three times the twice amount, the twice amount being what is needed in order to promote complete combustion.

02:18 So twice that would be 600, three times 600 would be 1 ,800 liter atmospheres per minute of oxygen are required.

02:31 But we're not providing oxygen, pure oxygen.

02:35 We're providing air, and air is only 21 % oxygen.

02:40 So how much air then do we need to provide per minute? we need to take our 1800 and divide it by the 0 .21%, to get our 8600 liter atmospheres per minute.

03:06 But then they threw a monkey wrench in.

03:10 The methane is being provided at 1 .50 atmospheres, but the air is being provided only at one atmosphere.

03:19 So in order to determine the volume, we simply need to divide by one atmosphere, and we get 8 ,600 liters per minute of air.

03:36 This is the first part of the question.

03:39 The second part is even more challenging.

03:42 It tells you that complete combustion does not occur, and that only 95 % of the carbon in methane becomes carbon dioxide, and the other 5 % of the carbon in methane becomes carbon monoxide.

03:59 So in one minute time, we'll have 200 liters of methane, and because of the stoichiometry being one -to -one for methane to carbon dioxide, we will then have 190 liters of carbon dioxide produced, that being 95 %, of the 100 liters of methane becoming, i'm sorry, the 200 liters of methane becoming 200 liters of carbon dioxide, but 95 % is only carbon dioxide, so that's now 190.

04:44 But that would be at 1 .5 atmospheres.

04:47 We want to figure out what the volume will be at one atmosphere...

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