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A large portion of metabolic energy arises from the biological combustion of glucose:
C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(g)
(a) If this reaction is carried out in an expandable container at 37°C and 780. torr, what volume of CO₂ is produced from 20.0 g of glucose and excess O₂?

Respuesta :

A large portion of metabolic energy arises from the biological combustion of glucose-

C6H12O6 (s) + 6O2 (g) ----> 6CO2 (g) + 6H2O (g)

Mass of glucose = 20g

Molar mass of glucose = 180.18 g/mol

=> Moles (glucose) = 20/180.18 = 0.111 mol

According to the equation,

Moles of CO2 = 6 × moles of glucose

=> Moles of CO2 = 0.666 mol

Also, we have

T = 310 K and P = 1.02 atm = 103351.5 pascals

Using the formula,

PV = nRT

=> 103351.5 x V = 0.666 x 8.314 x 310

=> V = (0.666 x 8.314 x 310)/103351.5 = 0.0166 m^3.

So, the volume of CO2 produced = 0.0166 m^3.

The products of combustion of glucose are water vapour and carbon dioxide.

The energy source is the simple sugar glucose (C6H12O6). Cellular respiration is the mechanism through which glucose is burned in bodily cells. Energy for living processes is provided by this combustion reaction.

he complete combustion of glucose will give carbon dioxide and water, therefore, the balanced chemical equation can be written as: C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(g)

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