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.

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

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