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