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Assume that an exhaled breath of air consists of \(74.8 \% \mathrm{~N}_{2}\), \(15.3 \% \mathrm{O}_{2}, 3.7 \% \mathrm{CO}_{2}\), and \(6.2 \%\) water vapor. (a) If the total pressure of the gases is \(0.980 \mathrm{~atm}\), calculate the partial pressure of each component of the mixture. (b) If the volume of the exhaled gas is \(455 \mathrm{~mL}\) and its temperature is \(37^{\circ} \mathrm{C}\), calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled. (c) How many grams of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would need to be metabolized to produce this quantity of \(\mathrm{CO}_{2}\) ? (The chemical reaction is the same as that for combustion of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). See Section 3.2.)

Step by step solution

01

To find the partial pressures of each component of the mixture, we will use the mole fraction of each component and multiply it by the total pressure of the mixture. For N鈧�: partial pressure = mole fraction * total pressure For O鈧�: partial pressure = mole fraction * total pressure For CO鈧�: partial pressure = mole fraction * total pressure For H鈧侽: partial pressure = mole fraction * total pressure #Step 2: Determine Mole Fractions for Each Gas#

We are given the percentages of each gas as a decimal: N鈧�: 74.8% O鈧�: 15.3% CO鈧�: 3.7% H鈧侽: 6.2% To convert these percentages into mole fractions, divide each percentage by 100: Mole fraction of N鈧� = 0.748 Mole fraction of O鈧� = 0.153 Mole fraction of CO鈧� = 0.037 Mole fraction of H鈧侽 = 0.062 #Step 3: Calculate and Display Partial Pressures of Each Gas#

02

Now we will use the mole fractions and the total pressure (0.980 atm) to find the partial pressures of each gas: Partial pressure of N鈧� = 0.748 * 0.980 atm = 0.733 atm Partial pressure of O鈧� = 0.153 * 0.980 atm = 0.150 atm Partial pressure of CO鈧� = 0.037 * 0.980 atm = 0.036 atm Partial pressure of H鈧侽 = 0.062 * 0.980 atm = 0.061 atm #Step 4: Calculate the Moles of CO鈧� Exhaled#

We know the volume (455 mL) and temperature (37掳C) of the exhaled gas. We also know the partial pressure of CO鈧� in the mixture. We can use the Ideal Gas Law (PV=nRT) to determine the number of moles of CO鈧�. First, convert the temperature to Kelvin: 37掳C + 273.15 = 310.15 K Next, convert the volume to liters: 455 mL / 1000 = 0.455 L Now, use the Ideal Gas Law for CO鈧�, with R = 0.0821 L*atm/mol*K: (0.036 atm) * (0.455 L) = n * (0.0821 L*atm/mol*K) * (310.15 K) #Step 5: Solve for the Number of Moles of CO鈧� Exhaled#

03

To find n (the number of moles of CO鈧�), divide both sides of the equation by (0.0821 L*atm/mol*K) * (310.15 K): n = (0.036 atm) * (0.455 L) / [(0.0821 L*atm/mol*K) * (310.15 K)] n = 0.0169 mol of CO鈧� exhaled #Step 6: Calculate the Grams of Glucose Needed to Produce CO鈧�#

To find the grams of glucose needed, first write down the balanced chemical equation for glucose's combustion: C鈧咹鈧佲倐O鈧� + 6 O鈧� 鈫� 6 CO鈧� + 6 H鈧侽 According to this equation, 1 mole of glucose produces 6 moles of CO鈧�, so to produce 0.0169 mol of CO鈧�, we would need (1/6) * 0.0169 mol of glucose. Now, calculate the molar mass of glucose (C鈧咹鈧佲倐O鈧�): (6*12.01) + (12*1.01) + (6*16.00) = 180.18 g/mol Finally, to find the grams of glucose needed, multiply the moles of glucose by its molar mass: (1/6) * 0.0169 mol * 180.18 g/mol = 0.0507 g of glucose needed to produce the given quantity of CO鈧�

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Pressure

When you're dealing with a mixture of gases, partial pressure is a handy concept. It's the pressure a gas would exert if it were alone in a container. In our problem, each gas in the exhaled breath contributes to the total pressure of 0.980 atm. The partial pressure can be figured out using the mole fraction of each gas.

Here's how you do it:

• First, calculate the mole fraction. The mole fraction is simply the percentage of each gas expressed as a decimal. For example, for nitrogen (\( \mathrm{N}_{2} \)), which is 74.8%, the mole fraction is 0.748.
• Next, multiply this mole fraction by the total pressure to get the partial pressure. So, for \( \mathrm{N}_{2} \), it would be \( 0.748 \times 0.980 = 0.733 \ \text{atm} \).

This method allows you to find the contribution of each gas to the overall pressure.

Mole Fractions

Mole fraction helps us understand the proportion of each component in a gas mixture. It鈥檚 an essential part of calculating partial pressures. The mole fraction can be determined using the following steps:

• Take the percentage of each gas component in the mixture. For example, oxygen (\( \mathrm{O}_{2} \)) is 15.3%.
• Convert this percentage to a decimal by dividing by 100. This gives us a mole fraction of \( 0.153 \) for \( \mathrm{O}_{2} \).

Mole fractions always add up to 1. This shows the entire composition of the mixture. Mole fractions are dimensionless and crucial in calculations related to partial pressures.

Combustion Reaction

Combustion reactions are processes where a substance combines with oxygen to produce heat and other products. In our exercise, glucose (\( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \)) undergoes a combustion reaction:

\[\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6} + 6 \mathrm{O}_{2} \rightarrow 6 \mathrm{CO}_{2} + 6 \mathrm{H}_{2}\mathrm{O}\]This equation shows that one molecule of glucose reacts with six molecules of oxygen, yielding six molecules of carbon dioxide and six molecules of water.

Combustion reactions are vital in many metabolic and industrial processes. They provide insights into how energy is released and utilized. Understanding the stoichiometry of these reactions helps in quantifying the reactants and products.

Chemical Stoichiometry

Chemical stoichiometry involves calculating the relationships between reactants and products in a chemical reaction. It鈥檚 like a recipe, telling us how much of each ingredient is needed.

Here's how it applies to our exercise:

• Look at the balanced combustion equation: \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6} + 6 \mathrm{O}_{2} \rightarrow 6 \mathrm{CO}_{2} + 6 \mathrm{H}_{2}\mathrm{O}\).
• This tells us that one mole of glucose produces six moles of \( \mathrm{CO}_{2} \).
• If we have \( 0.0169 \ \text{mol} \) of \( \mathrm{CO}_{2} \), we can use stoichiometry to find that only one-sixth of that (\( 0.00282 \ \text{mol} \)) amount of glucose reacted.
• Calculate the mass of this glucose: \( 0.00282 \ \text{mol} \times 180.18 \ \text{g/mol} = 0.0507 \ \text{g} \).

Stoichiometry is essential for determining how much of a chemical is consumed or produced. It ensures balanced reactions and accurate scientific calculations.