Question

Calculate the volume of dry CO2 produced at body temperature (37 ∘C) and 0.980 atm when 24.0 g of glucose is consumed in this reaction.

Answer 1

Molar mass of C6H12O6,

MM = 6*MM(C) + 12*MM(H) + 6*MM(O)

= 6*12.01 + 12*1.008 + 6*16.0

= 180.156 g/mol

mass of C6H12O6 = 24 g

mol of C6H12O6 = (mass)/(molar mass)

= 24/1.802*10^2

= 0.1332 mol

Balanced chemical equation is:

C6H12O6 + 6O2 —> 6CO2 + 6 H2O

According to balanced equation

mol of CO2 formed = (6/1)* moles of C6H12O6

= (6/1)*0.1332

= 0.7993 mol

Given:

P = 0.98 atm

n = 0.7993 mol

T = 37.0 oC

= (37.0+273) K

= 310 K

use:

P * V = n*R*T

0.98 atm * V = 0.7993 mol* 0.08206 atm.L/mol.K * 310 K

V = 20.748 L

Answer: 20.7 L