C6h12o6(s) + 6o2(g) --> 6h2o(g) + 6co2(g) how many liters of co2 are produced when you start with 13.2 grams of glucose (c6h12o6) in excess oxygen at stp? 0.439 liters co2 10.0 liters co2 13.5 liters co2 20.4 liters co2

One
Balance the Equation.
This has been done for you or it is given. Anyway this step is finished, but it must always be done.

Two
Find the molar mass of C6H12O6
6C = 6 * 12 = 72
12H = 12*1 = 12
CO = 6 * 16 = 96
Mol Mass = 180 grams / mol

Three
Find the mols of C6H12O6
n = ???
Molar Mass = 180 grams / mol
given mass = 13.2 grams.

n = given mass / molar mass
n = 13.2 / 180
n = 0.07333333 mols.

Four
Find the mols CO2
1 mol C6H12O6 will produce 6 mols CO2
0.0733333 mols will produce x

1/0.073333 = 6/x Cross multiply
x = 0.073333 * 6
x = 0.4399 moles.

Five
Find the volume given the conditions for temperature and pressure are STP conditions.
PV = nRT
R = 0.082057 
n = 0.43999
P = 1 atmosphere
T = 0 + 273 = 273
V = ???
1 * V = 0.43999 * 082057 * 273
V = 10.2 L
B

Note if you give out awards I'd sure appreciate one. This was a lot of typing