8. GAS LAW OXIDATION OF GLUCOSE (C6H12O6) BY THE BODYPRODUCES CARBON DIOXIDE WHI

Question

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

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