Question

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.

Answer 1

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

Answer 2

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