Calculate the volume of dry co2 produced at body temperature (37 ∘c) and 0.970 atm when 25.5 g of glucose is consumed in this reaction.

I believe the balanced chemical equation is:

C6H12O6 (aq) + 6O2(g) > 6CO2(g) + 6H2O(l) 

First calculate the moles of CO2 produced:

moles CO2 = 25.5 g C6H12O6 * (1 mol C6H12O6 / 180.15 g) * (6 mol CO2 / 1 mol C6H12O6)

moles CO2 = 0.8493 mol

Using PV = nRT from the ideal gas law:

V = nRT  / P

V = 0.8493 mol * 0.08205746 L atm / mol K * (37 + 273.15 K) / 0.970 atm

V = 22.28 L

The volume of the dry CO₂ produced is 22.30 liters.

The given parameters;

The balanced chemical reaction of the given compounds;

Molecular mass of glucose ( = 180 g/mol)

180 g of glucose ---------------- 6 moles of CO₂

25.5 g of glucose ---------------- ? moles of  CO₂

The volume of the  dry CO₂ produced is calculated form Ideal gas law;

PV = nRT

where;

R is the ideal gas constant = 0.082057 L. atm /mol K

Thus, the volume of the dry CO₂ produced is 22.30 liters.

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yes, water is wet, just as it makes other things wet, it is wet itself, kinda like ice, when ice is in water, the ice makes water cold, and ice is cold, therefore with that example, water is wet

what is the question here. i think you messed up.