Volume of methane from the decomposition of 1 kg of glucose

The bacterial decomposition of glucose yields methane and CO2. what would the volume of methane be from the decomposition of 1 kg of glucose at 300 degrees k and 825 torr? show balanced reaction.

The balance i got was C6H12O6 ------> 3CH4 + 3O2. then I got 1.086 atm for P (825-760). Next I got 5.56 mol for glucose from 1kg(1000g/1kg) / 180g-MW of gulcose. I wasnt sure what to do from here and assumed I would do 5.56 mol glucose (3CH4/1mol glucose) = 16.68 mol CH4. Then I used ideal gas law formula: V=nRT/P... (16.68 x .082 x 300)/1.086 atm = 377.8 about 378L. I know this is wrong.. Can you please tell me where I went wrong, and show me what I should do.

Thank you so much for helping...

Your balanced equation should read : C6H12O6 ------> 3CH4 + 3CO2

1000 g C6H12O6 x (1 mole C6H12O6 / 180 g C6H12O6) x (3 moles CH4 / 1 mole C6H12O6) = 16.7 moles CH4. Likewise, 16.7 moles of CO2 are produced. So the mole fraction of each gas = 0.500.
P CH4 = P total x mole fraction CH4 = 825 torr x 0.50 = 413 torr

PV = nRT

P = CH4 pressure in atm = 413 torr x (1 atm / 760 torr) = 0.543 atm
V = CH4 volume in L = ?
n = moles of CH4 = 16.7
R = gas constant = 0.0821 L atm / K mole
T = Kelvin temperature = 300 K

V = nRT / P = (16.7)(0.0821)(300) / (0.543) = 757 L CH4