When eaten, dietary carbohydrates are digested to yield glu-cose (C6H12O6), which is then metabolized to yield carbon dioxide and water:
Balance the equation, and calculate both the mass in grams and the volume in liters of the CO2 produced from 66.3 g of glucose, assuming that 1 mol of CO2 has a volume of 25.4 L at normal body temperature.

Question

When eaten, dietary carbohydrates are digested to yield glu-cose (C6H12O6), which is then metabolized to yield carbon dioxide and water: 
Balance the equation, and calculate both the mass in grams and the volume in liters of the CO2 produced from 66.3 g of glucose, assuming that 1 mol of CO2 has a volume of 25.4 L at normal body temperature.

Accepted Answer

welcome back everyone. We're told that fermentation involves converting glucose into ethanol and carbon dioxide, calculate the mass and grams and volume. And leaders of ethanol that is produced by the fermentation of 105.3 g of glucose assumed that at 25 degrees Celsius, one mole of ethanol has a volume of 58.39 mL. Our first step is to write out a balanced equation beginning with our conversion of glucose. So C six H 12 06 and this solid is being converted through fermentation into ethanol. So we have C two H five oh H. And we're also going to produce carbon dioxide as our second product. Our first step is to make sure that our equation is balanced. So we have carbon, hydrogen and oxygen on both sides of our equation, counting our atoms on the reactant side, we have six moles of our carbon atoms. We have 12 moles of hydrogen on the reactant side. And then for oxygen we have six on the reactant side on our product side, we have a total of two, Three moles of our carbon on the product side, we have 56 moles of our hydrogen on the product side, sorry. And then we have 23 moles of oxygen on the product side. So now we need to bounce things out. So we're going to begin by placing a coefficient of two in front of our ethanol on the product side, which is going to give us now a total of 45 carbon atoms on the product side that we can actually mend that to a six if we place a coefficient of two in front of carbon dioxide as well. So this is going to now give us 456 carbon atoms For hydrogen. We can count a total of five times two, which would give us 10 and then we would have 11 And then 12. Since we're multiplying this hydrogen by two. So this would go to 12. And now we have for our oxygen two times one which is two plus our four oxygen's here, which is going to give us six oxygen's on the product side. And now everything is balanced in our equation. So now that we have our balanced equation, we want to make note of our molar mass of our ethanol and our glucose. So beginning with our glucose. And let's actually so C six H 06 referring to its molar mass. Using our periodic table, we're going to calculate a total of 180. g per mole. And then for our ethanol C two H five oh H. We're going to calculate a molar mass from our periodic table of 46.7 g per mole. So for our first part of our answer, finding our massive ethanol needed to produce or that will be produced by our fermentation of our glucose. We're going to begin with the information that we know right now which is our mass of our glucose given in the prompt as 105.3 g of our glucose. So now we want to cancel out grams of our glucose. So we're going to go from grams of glucose in the denominator to molds of glucose in the numerator where from our periodic table we will utilize the molar mass of glucose which we see as 180.6 g for one mole. Now canceling out our units of grams of glucose. We can go from moles of glucose to our moles of ethanol. So moles of glucose is in the denominator and moles of ethanol is going to be in our numerator where we would get this ratio from our balanced equation. So looking at our balanced equation, we have two moles of ethanol for one mole of our glucose. And so we're going to plug that molar ratio in two moles of ethanol for one mole of glucose. We can now cancel out moles of glucose and focus on going from moles of our ethanol in the next line to so for moles of ethanol in the denominator to our mass of ethanol that is produced in the fermentation of our glucose. So grams of C two H five oh H where we are going to plug in our molar mass of our ethanol as 46 point oh seven g for one more. So canceling out our moles of ethanol, we're left with grams of our ethanol And we're going to have a mass produced equal to the value of 53.88 g. So this is going to be our first answer as the mass of our ethanol produced from the fermentation. And now we just need the volume of our ethanol produced from the fermentation of our glucose. So that's going to be part two from fermentation. So beginning the same way with our known mass of our glucose. We have 105.3 g of our glucose given from the prompt. And this is going to be multiplied again by glucose is molar mass to get rid of grams of glucose. So in the denominator we want glucose C six H 12 06. Specifically 180.06 equivalent to one mole of our glucose. So this is again our molar mass from the periodic table. We can now cancel out grams of glucose moving into in our denominator moles of glucose And in our numerator. And sorry, that's C6 H 1206. So just to make sure that this is visible, let's scoot this over a bit. So in our numerator we want moles of our ethanol C two H five oh H. We're going to utilize that multiple ratio which we determined from our previous step and from our bounced equation is a 1 to 2 molar ratio allowing us to get rid of moles of glucose and now focus on going from moles of our ethanol. H50 H two moles of our or rather grams of our ethanol. Sorry about that in our numerator, plugging in our molar mass of ethanol from our periodic tables as we did before. We have a mass of 46 point oh seven g for one mole of ethanol, canceling out moles of ethanol. We're left with grams and we need to get to leaders as our volume. And so we're going to multiply by our next conversion factor where correction. We're actually going to not utilize the molar mass but instead we can directly go from moles in the denominator of ethanol to our next unit which is leaders of ethanol. And this is because if we go back up to our prompt, We'll see that they told us that we're assuming we have one mole of ethanol according to the volume of 58.39 ml. So we actually need to plug in in the numerator ml And that is 58.39 ml of our ethanol for one mole of our ethanol. And so now we can cancel out moles of ethanol. And now we need leaders as our final unit for volume. So recall that we need middle leaders in the denominator leaders in the numerator and we recognize our prefix milli tells us we have 10 to the negative third power of our base unit. Leaders now canceling out milliliters of ethanol. We are left with leaders of our ethanol produced from the fermentation of glucose as our final unit. And this is going to yield a result of our volume of ethanol that is produced being of 0.068- leaders. And this would be our second and final answer as our volume of ethanol that is produced from our fermentation of glucose. So it's highlighted in yellow are two final answers of the mass and volume of our ethanol that we produced from glucose being fermented. I hope everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

Verified Solution

Key Concepts

Stoichiometry

Balancing Chemical Equations

Gas Volume and Molar Relationships