The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

C6H12O6(aq) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
Calculate the volume of dry CO2 produced at body temperature (37°C) and 0.980 atm when 25.5 g of glucose is consumed in this reaction.

mols glucose = 25.5g/molar mass glucose.

mols CO2 produced = 6x that (Look at the coefficients in the balanced equation.)
Use PV = nRT and solve for volume at the conditions listed. Remember T must be in kelvin.

To calculate the volume of dry CO2 produced at body temperature (37°C) and 0.980 atm when 25.5 g of glucose is consumed, we need to use the ideal gas law.

The ideal gas law states: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

First, we need to find the number of moles of glucose (C6H12O6). To do this, we can use the molar mass of glucose (180.16 g/mol):

Moles of glucose = mass of glucose / molar mass of glucose
Moles of glucose = 25.5 g / 180.16 g/mol
Moles of glucose ≈ 0.1416 mol

According to the balanced equation, 1 mole of glucose reacts to produce 6 moles of CO2. Thus, we can determine the number of moles of CO2 produced:

Moles of CO2 = moles of glucose × 6
Moles of CO2 = 0.1416 mol × 6
Moles of CO2 ≈ 0.8496 mol

Next, we need to convert the temperature to Kelvin:

T(K) = 37°C + 273.15
T(K) ≈ 310.15 K

Now we can use the ideal gas law to find the volume of CO2:

PV = nRT

V = (nRT) / P
V = (0.8496 mol × 0.0821 L·atm/mol·K × 310.15 K) / 0.980 atm
V ≈ 21.074 L

Therefore, the volume of dry CO2 produced is approximately 21.074 liters at body temperature (37°C) and 0.980 atm when 25.5 g of glucose is consumed.

To calculate the volume of dry CO2 produced in the given reaction, we need to use the ideal gas law equation:

PV = nRT

Where:
P is the pressure of the gas (in atm),
V is the volume of the gas (in liters),
n is the number of moles of gas,
R is the ideal gas constant (0.0821 L.atm/mol.K), and
T is the temperature (in Kelvin).

First, we need to find the number of moles of CO2 produced.

Step 1: Calculate the number of moles of glucose (C6H12O6).
To do this, we use the formula:

moles = mass / molar mass

The molar mass of glucose (C6H12O6) can be calculated by adding up the atomic masses of its constituent elements.

molar mass of C6H12O6 = (6*12.01) + (12*1.01) + (6*16.00)
= 180.18 g/mol

moles of glucose = 25.5 g / 180.18 g/mol
= 0.1417 mol

Step 2: Calculate the number of moles of CO2 produced.
Based on the balanced equation, for every one mole of glucose consumed, six moles of CO2 are produced.

moles of CO2 = 0.1417 mol * 6
= 0.8502 mol

Step 3: Convert the temperature to Kelvin.
Given that the temperature is 37°C, we need to convert it to Kelvin by adding 273.15 to it.

T = 37°C + 273.15
= 310.15 K

Now, we have all the necessary values to calculate the volume of dry CO2 using the ideal gas law.

PV = nRT

V = nRT / P
= (0.8502 mol * 0.0821 L.atm/mol.K * 310.15 K) / 0.980 atm
= 21.93 L

Therefore, the volume of dry CO2 produced at body temperature (37°C) and 0.980 atm when 25.5 g of glucose is consumed is approximately 21.93 liters.