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.

Using the equation PV = nRT
Therefore; V = nRT / P
Need moles of glucose converted to moles of the product gas (CO2).

Molecular weight calculation:
C 6 X 12.01 = 72.06
H 12 X 1.01= 12.12
O 6 X 16.00 = 96.00
 sum = 180.18
25.5 g of C6H12O6 ( 1 mol C6H12O6 / 180.18 g) ( 6 mol CO2 / 1 mol C6H12O6) = 0.84915 mol CO2 gas.
Convert temp: 37 °C + 273.15 = 310.15 K
V= ((0.84915 mol)× (0.0821 L atm / mol K) (310.15 K))/0.980 atm
 V = 22.0635 L
  = 22.06 L CO2

  The  volume  of dry  gas   is 22.11 L

   calculation
by  use  of ideal gas  equation that is Pv=nRT
where  P (pressure) =  0.980  atm
            V= ?
            n =  number of moles
            R (gas constant)  =  0.08205  L.atm/Mol.K
            T(Temperature)  = 37 +273  = 310 k

find  n (number of moles)
write  the  equation for reaction

C6H12O6  +6O2  = 6CO2 +6H2O

find the moles  of C6H12O6  =  moles/molar  mass  = 25.5/180 = 0.142  moles

by use of mole ratio  between  C6H12O6 to CO2  which is   1:6  the  moles of CO2 =  0.142 x6 =  0.852  moles  therefore n=  0.852 moles

by  making  V the formula of the  of the subject   V= nRT/P

V =( 0.852moles  x0.08205 L.atm/mol.K x310 k)  / 0.980 atm  =  22.11 l

Ithink the answer is c i hope it my friend message me if i’m wrong and i’ll change my answer and fix it for you