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barnuts

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

onyebuchinnaji

The volume of the dry CO₂ produced is 22.30 liters.

The given parameters;

  • temperature of the body, T = 37 ° C = (273 + 37) K = 310 K
  • pressure of the body, P = 0.97 atm

The balanced chemical reaction of the given compounds;

[tex]C_6H_{12}O_6 \ +\ 6O_2 \ ----> \ 6CO_2\ + \ 6H_2O[/tex]

Molecular mass of glucose ([tex]C_6H_1_2O_6[/tex] = 180 g/mol)

180 g of glucose ---------------- 6 moles of CO₂

25.5 g of glucose ---------------- ? moles of  CO₂

[tex]= \frac{25.5 \times 6}{180} \\\\= 0.85 \ mol \ of \ CO_2[/tex]

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

[tex]V = \frac{nRT}{P} \\\\V = \frac{0.85 \times 0.082057\times 310}{0.97} \\\\V = 22.30 \ liters[/tex]

Thus, the volume of the dry CO₂ produced is 22.30 liters.

Learn more here: https://brainly.com/question/24542515

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