Assume that an exhaled breath of air consists of 74.8% N 2 , 15.3% O 2 , 3.7% CO 2 , and 6.2% water vapor. (c) How many grams of glucose (C 6 H 12 O 6 ) would need to be metabolized to produce this quantity of CO 2 ? (The chemical reaction is the same as that for combustion of C 6 H 12 O 6 . See Section 3.2 and Problem 10.57.)

Hi everyone for this problem, we're told that a 1.506 reaction flask contains 14.22 of oxygen gas, 61.27% carbon dioxide and 24.51% water vapor at 30 degrees Celsius, the total pressure inside the flask is 24.61 kill a pascal's. We need to calculate the mass of nap filling that was burned in order to produce this amount of water vapor and were given the chemical equation. Okay, so for this problem, we need the ideal gas law and the ideal gas law tells us that P. V. Is equal to N. R. T. And for us to calculate the mass of naphthalene, we first need to figure out our amount of water vapor and so we can rearrange our gas law to solve for moles of water vapor and then from moles of water vapor, we can use our multiple ratio to go from moles of water vapor to mass of nap filling. So let's go ahead and rearrange this so that we're solving for moles. And when we do we get that N. Is going to equal P. V over R. T. So first we want most of water vapor. Okay, So let's write out what we know and are given so that we can solve for this. Okay, so p is our pressure and we know that when we're dealing with gasses. So I'll ride over here given So the first thing we're gonna solve for is P. And four gasses, the pressure of a gas is going to equal the mole fraction of that gas times the total pressure. Okay. And so because we're dealing with water vapor, they tell us that our water vapor is .51%. If we convert this percent to a decimal, that is going to give us our mole fraction of water vapor. Okay, so The pressure of our water vapor is the Mole fraction, which is the 24.51%. So we'll say that is 0.24 times the total pressure. They tell us that our total pressure And the problem is 24.61 kill a pascal's. Okay? So when we solve this we get 6.319 killer pascal's. But we need to convert this killer pascal's to A. T. M. So six point 19 Killer Pascal's and one killer pascal. There is 1000 pascal's And in one a. t. m. There is 100 and 1325 pascal's. Okay, so let's make sure our units cancel here. Killer pascal's cancel pascal's cancel. We're left with A T. M. So our Pressure of water vapor and ATM is 0.05 A T. M. Okay, so we have pressure. So let's go ahead and write that here. So n. is equal to pressure we just saw for pressure. 0.5953 A. T. M. Our volume We're told is 1.50 L yep, a reaction flask It says is 1.50 L. So 1.50 L. So pressure times volume over our is our gas constant. This isn't a value that's going to be given. It's something that we should know. And that value is 0. 8206. Are leader times atmosphere over more times kelvin. Okay, so that's our and then our temperature, We're told it's 30°C. However, we need to convert this to kelvin because our gas constant, if you look at the units, its leaders per atmosphere Over most times leaders times atmosphere over most times Kelvin. So we need our temperature in Kelvin. So we add 273.15 Kelvin and that gives us 303 . Kelvin. Okay, so we have everything in our problem to solve from moles of water vapor. So our moles of water vapor is equal to 3. times 10 to the -3. And that's most of water vapor. Okay, so now that we have our molds of water vapor, we can go to grams of naphthalene, which is what we're looking for. We're looking for a massive nap feelin. So let me make some space here going to make this smaller. Okay, so our moles of water vapor, we need to go from moles of water vapor to grams of naphthalene. Looking at our reaction, we see that for every formals of water vapor we have one mole of nap filling. So let's go ahead and write that out. And one more of naps Celine for every one mole of naps Selene, we have four moles of water vapor. Okay? And we want to calculate massive naphthalene. So here we see that our moles of water vapor cancel. And we're left with moles of naphthalene. And in order to find out our massive naphthalene and grams, we need the molar mass of naphthalene. So in one mole of naps Celine using our periodic table to calculate its smaller mass, We find that our molar mass of Naphthalene is 128. g of naps Selene. And now our molds of naphthalene cancel. And we're left with grams of naphthalene. And so now We do this calculation and we get 0. grams of naps doing which is our final answer. Okay, That is the end of this problem. And I hope this was helpful

Ch.10 - Gases

Brown - Chemistry: The Central Science 14th Edition

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Ch.10 - Gases

Problem 107c

Chapter 10, Problem 107c

Assume that an exhaled breath of air consists of 74.8% N₂, 15.3% O₂, 3.7% CO₂, and 6.2% water vapor. (c) How many grams of glucose (C₆H₁₂O₆) would need to be metabolized to produce this quantity of CO₂? (The chemical reaction is the same as that for combustion of C₆H₁₂O₆. See Section 3.2 and Problem 10.57.)

Verified step by step guidance

Determine the total volume or mass of the exhaled breath to find the mass of CO₂ it contains. Assume a specific volume or use a typical value for a breath.

Calculate the mass of CO₂ in the exhaled breath using its percentage composition. Convert the percentage to a mass fraction and multiply by the total mass or volume of the breath.

Write the balanced chemical equation for the combustion of glucose, C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O.

Use stoichiometry to relate the moles of CO₂ produced to the moles of glucose consumed. From the balanced equation, 1 mole of glucose produces 6 moles of CO₂.

Calculate the mass of glucose needed by converting the moles of glucose to grams using the molar mass of glucose (180.16 g/mol).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows us to calculate the amounts of substances consumed and produced in a reaction based on balanced chemical equations. In this context, understanding stoichiometry is essential to determine how many grams of glucose are needed to produce a specific amount of carbon dioxide.

Combustion of Glucose

The combustion of glucose (C6H12O6) is a biochemical process where glucose reacts with oxygen to produce carbon dioxide and water, releasing energy. The balanced equation for this reaction is C6H12O6 + 6 O2 → 6 CO2 + 6 H2O. Knowing this reaction is crucial for calculating the amount of glucose required to generate a given quantity of CO2, as it directly relates the moles of glucose to the moles of CO2 produced.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is essential for converting between the mass of a substance and the number of moles, which is necessary for stoichiometric calculations. In this problem, calculating the molar mass of glucose will enable us to convert the moles of glucose needed to produce the specified amount of CO2 into grams.