How many liters of  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? a)0.439 liters co2 b)10.0 liters co2 c)13.5 liters co2 d)20.4 liters co2

The balanced equation for the above reaction is as follows
C₆H₁₂O₆(s) + 6O₂(g) --> 6H₂O(g) + 6CO₂(g)
the limiting reactant in the equation is glucose as the whole amount of glucose is used up in the reaction.  
the amount of C₆H₁₂O₆ used up - 13.2 g
the number of moles reacted - 13.2 g/ 180 g/mol  = 0.073 mol
stoichiometry of glucose to CO₂ - 1:6
then number of CO₂ moles are - 0.073 mol x 6 = 0.44 mol
As mentioned this reaction takes place at standard temperature and pressure conditions, 
At STP 1 mol of any gas occupies 22.4 L
Therefore 0.44 mol of CO₂ occupies 22.4 L/mol  x 0.44 mol = 9.8 rounded off - 10.0 L
Answer is B) 10.0 L CO₂

answer: mole of the gas

answer: mole of the gasexplanation:

answer: mole of the gasexplanation: this is because 1 mole of a gas is 22.4dm³, which is use in ideal gas