A large portion of metabolic energy arises from the biological combustion of glucose:
C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(g)
(b) If the reaction is carried out at the same conditions with the stoichiometric amount of O₂, what is the partial pressure of each gas when the reaction is 50% complete (10.0 g of glucose remains)?

for this question, we are going to be using this bounced equation for the combustion of glucose. And for the first part of this question, they are wanting us to find find the volume of CO two produced by 20 g of glucose. So they tell us that our mass of glucose is equal to 20 g, and they tell us that the temperature for the reaction is 37 degrees Celsius. I'm just gonna add 273 Calvin to this.

So our temperatures going to be 3 10 Calvin, they tell us that the pressure for the reaction is equal to 780 toured. I'm just gonna change in the atmosphere by times hand by 0.0 131579 atmospheres per tour. So our pressure is gonna be 1.3 atmospheres and they're wanting us to find the volume of CO two for this reaction, right.

So for part A. To find out the volume of co two, we need to find out the moles of glucose to start out with so moles of glucose. And we know that the molar mass abbreviated mm of glucose is equal to 180.15 g per No. So to get our molds, we're just gonna take the amount of glucose we have in our reaction.

So 20 g of glucose I was gonna divide that by its Moeller Mass. So our moles of glucose is gonna equals 0.111 moves. All right, so now that we have our moles of glucose, we can now get our moles of co two. So if we go up here to our balanced equation, we can see that we have six moles of CO two for every one mole of our glucose.

So we're gonna take the moles of glucose. So 0.111 Mose glucose And we're in a times that by six moles of CO two and divide that for every one mole of glucose. Okay, so our moles of co two is equal to 0.666 MOCs. All right, so now we just need to find the volume of co two So volume is equal to moles. Times are our value. Times are temperature divided by our pressure. So our volume is equal to our moles of co two So 0.666 Mose times that by our our value, which is 0.8 to 1 leaders atmosphere per mole. Calvin and I'm a times that by our temperature. So 3. 10 Calvin and we're gonna divide that by our pressure. So is 1.3 atmospheres. So our volume of CO two is gonna equal 16.5 leaders.

All right, so that is how we solve the first part of this question. For the second part of this question, they are asking us to find the partial pressures of each gas when the reaction has only been 50% completed. So when our reaction is only at 50% completed, we have half of our 02 molds consumed by the reaction.

So if we have one more of 02 every one mole of CO two based on our kept balanced chemical equation for our moles of 02 co two in each 20 are all gonna equals 0.333 moles because it's half of what it was when the reaction was completed. 100%.

So for the partial pressure of water, it says that we are at. We know that for the pressure for water at 37 degrees C or 3. 10 Calvin are partial. Pressure of water is equal to 48.8 tour. So I'm just gonna convert this into atmospheres by times ing it by 0.0 131579 atmospheres per tour. S

So are partial. Pressure for water is gonna be 0.642 atmospheres. All right. And this is our partial pressure for the water vapor, and they tell us that the total pressure is equal to 1.3 atmospheres. So if we know that the partial pressure for co two and the partial pressure for oxygen are equal because they have the same number of moles, we can say that the total pressure minus the partial pressure of water, is equal to double that of the pressure for co two and oxygen.

So I calculate this out is gonna be 0.9 658 atmospheres again, This is both for co two and oxygen. So to get the partial pressure First co two and oxygen, we're just gonna divide this number by two 9.46 There we go. Zero point 9658 divided by to So this is gonna equals zero point for 8 to 9 atmospheres.

So this is gonna be the partial pressure for co two and the partial pressure for oxygen. And that is how we calculate the partial pressures for this reaction.

Learn more about partial pressure https://brainly.com/question/13199169

#SPJ4