8. GAS LAW OXIDATION OF GLUCOSE (C6H12O6) BY THE BODYPRODUCES CARBON DIOXIDE WHICH IS EXPELLED THROUGH THE LUNGS. C6H12O6+6O2(G)----6CO2(G)+ 6 H2O (L) CALCULATE THE VALUME OF CO2 PRODUCE AT BODYTEMPERATURE (37 DEGREES CELSIUS) AND 0.970 ATM WHEN 24.5 G OFGLUCOSE IS CONSUMED IN THIS REACTION 8. GAS LAW OXIDATION OF GLUCOSE (C6H12O6) BY THE BODYPRODUCES CARBON DIOXIDE WHICH IS EXPELLED THROUGH THE LUNGS. C6H12O6+6O2(G)----6CO2(G)+ 6 H2O (L) CALCULATE THE VALUME OF CO2 PRODUCE AT BODYTEMPERATURE (37 DEGREES CELSIUS) AND 0.970 ATM WHEN 24.5 G OFGLUCOSE IS CONSUMED IN THIS REACTION OXIDATION OF GLUCOSE (C6H12O6) BY THE BODYPRODUCES CARBON DIOXIDE WHICH IS EXPELLED THROUGH THE LUNGS. C6H12O6+6O2(G)----6CO2(G)+ 6 H2O (L) CALCULATE THE VALUME OF CO2 PRODUCE AT BODYTEMPERATURE (37 DEGREES CELSIUS) AND 0.970 ATM WHEN 24.5 G OFGLUCOSE IS CONSUMED IN THIS REACTION
Explanation / Answer
We Know that : The given Equation is : C6H12O6+6O2(G)----6CO2(G) + 6H2O (L) According to ideal gas equation: PV = n RT 0.970 atm x V = 24.5g / 180 g / mol x 0.0821 atm-L / mol-K x 310K V = 3.571 L Volume of theCO2 = Volume of the glucose / 6 From the above balanced equation. = 3.571 L / 6 = 0.5952 L