Gas Law/Balancing Reactions question?

Answer is A but unsure why

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  • 8 months ago
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    (0.054 g C6H12O6) / (180.1559 g C6H12O6/mol) = 0.00029974 mol C6H12O6

    n = PV / RT = (1.00 atm) x (0.250 L) / ((0.082057366 L atm/K mol) x (295 K)) = 0.01033 mol O2

    0.00029974 mole of C6H12O6 would react completely with 0.00029974 x (6/1) = 0.00179844 mole of O2, but there is more O2 present than that, so O2 is in excess and C6H12O6 is the limiting reactant.

    (0.00029974 mol C6H12O6) x (6 mol CO2 / 1 mol C6H12O6) =

    0.00179844 mol CO2

    (0.00029974 mol C6H12O6) x (6 mol H2O / 1 mol C6H12O6) =

    0.00179844 mol H2O

    (0.01033 mol O2 initially) - (0.00179844 mol O2 reacted) =

    0.00853156 mol O2 left over

    (0.00179844 mol CO2) / (0.00179844 mol + 0.00179844 mol + 0.00853156 mol) =

    0.148 (the mole fraction of CO2)

    So answer a.)

  • 8 months ago

    The 0.054g of glucose is 

    (0.054g / (180 g/mol)) = 0.300 millimoles.

    The 250 mL of O2 at 1.00 atm and 295K is

    (0.25 L)*(1 mol/22.4 L)*(273K/295K) = 10.33 millimoles.

    The reaction uses up all the glucose, but only 1.8 millimoles of the O2, so the gas now in the container is 1.8 millimoles of CO2, 1.8 millimoles of H2O, and  8.53 millimoles of O2.

    Finally, 1.8/(1.8 + 1.8 + 8.53) = 0.148

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