The metabolic breakdown of glucose (C 6 H 12 O 6 ) (MW:180.156 g/mol) is given by the following equation: C 6 H 12 O 6 (s) + 6 O 2 (g) → 6 CO 2 (g) + 6 H 2 O (l) Calculate the volume (in mL) of CO 2 produced at 34°C and 1728.9 torr when 231.88 g glucose is used up in the reaction.

Here we're told that the metabolic breakdown of glucose, which is C6H1206 with a molecular weight of this, is given by the following equation. Now here it says calculate the volume in milliliters of CO2 produced at 34 degrees Celsius and 1728.9 torrs when 231.88 grams of glucose is used up in the reaction. Alright. So CO 2 is a gas, we're looking for its volume. Remember, ideal gas law is p v equals n r t. We want just the volume, so volume is equal to n r t over p. We know what r is, it's a gas constant. So plug that in. Temperature, you add 273.15 to this. That gives us 307.15 Kelvin. Pressure, remember we have 1728.9 Torr. One atmosphere is equal to 760 Torr. So that comes out to 2.2 749 atmospheres. Now we're almost done, but we can't find the volume just yet because we're still missing our moles of our gas. How do we determine this portion? Well, they give me the grant of glucose. If we use stoichiometry, we can find the moles of glucose. And if we know the moles of glucose, we know the moles of everyone within the equation by doing a mole to mole comparison. So at this point, we're gonna do some stoichiometry to help us. We have 231.88 grams of glucose. Now remember, one mole of glucose is equal to 180.156 grams. So grams cancel out. And according to my balanced equation, for every one mole of glucose, there's 6 moles of c o 2. K. So that's our mole to mole comparison. Moles of CO 2. So when we do that, we're going to get 7.72 to 6 moles of CO2. And that's what's going to go here. Moles cancel out, Kelvins cancel out, atmospheres cancel out. So at the end, we have liters. So this comes out to be 85.56 liters. But remember, we don't want liters here, we want milliliters. So one more conversion. One milliliter is equal to 10 to the negative 3 liters. Alright. So this is gonna come out to be 0.08556 liters. In terms of sig figs, 34 has 2 sig figs. We're just gonna do this as 0.086 oops. Yeah. 6. Sorry. Actually we need to convert this. This is liters, we have to flip everything. We want milliliters, So it's 10 to the negative 3 here for every 1 milliliter. Liters cancel out, and now we're gonna have milliliters. Alright. So then this is going to come out to be 85,560 milliliters. We want 2 sig figs. So just make this 8.6 times 104ml. This would be our final answer for this particular question.

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7. Gases

Gas Stoichiometry

General Chemistry

7. Gases

Gas Stoichiometry

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Multiple Choice

The metabolic breakdown of glucose (C₆H₁₂O₆) (MW:180.156 g/mol) is given by the following equation:
C₆H₁₂O₆ (s) + 6 O₂ (g) → 6 CO₂ (g) + 6 H₂O (l)
Calculate the volume (in mL) of CO₂ produced at 34°C and 1728.9 torr when 231.88 g glucose is used up in the reaction.

8.6×10⁵ mL

6.8×10⁴ mL

8.6×10⁴ mL

4.3×10⁴ mL

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Verified step by step guidance

Step 1: Begin by calculating the number of moles of glucose (C6H12O6) used in the reaction. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Here, the mass of glucose is 231.88 g and the molar mass is 180.156 g/mol.

Step 2: Use the stoichiometry of the balanced chemical equation to determine the moles of CO2 produced. According to the equation, 1 mole of glucose produces 6 moles of CO2. Multiply the moles of glucose by 6 to find the moles of CO2.

Step 3: Convert the temperature from Celsius to Kelvin using the formula: \( T(K) = T(°C) + 273.15 \). This is necessary for using the ideal gas law.

Step 4: Convert the pressure from torr to atm using the conversion factor: \( 1 \text{ atm} = 760 \text{ torr} \). This will allow you to use the ideal gas law.

Step 5: Apply the ideal gas law \( PV = nRT \) to find the volume of CO2 produced. Rearrange the formula to solve for volume \( V = \frac{nRT}{P} \). Use \( R = 0.0821 \text{ L atm/mol K} \) as the ideal gas constant. Finally, convert the volume from liters to milliliters by multiplying by 1000.