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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an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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