Question #a6390

The maximum mass of water that can be produced from your reaction is #"32.9." mu"g"#.

So, you're dealing with the combustion reaction of glucose, #C_6H_12O_6#; the balanced chemical equation for this reaction looks like this

#C_6H_12O_6 + color(red)(6)O_2 -> 6CO_2 + color(red)(6)H_2O#

Notice the #1:color(red)(6)# mole ratio that exists between both glucose and oxygen, and between glucose and water.

This means that, regardless of how many moles of glucose react, you'll always need 6 times more moles of oxygen and produce 6 times more moles of water.

So, the first thing to do is determine whether or not you have a limiting reagent, i.e. if one reactant is in insufficient amount compared with the other.

Use glucose and oxygen's molar masses to determine how many moles of each you have

#"55.0"mu"g" * "1 mole glucose"/"180.16 g" = "0.3053 "mu"moles glucose"#

#"35.0"mu"g" * "1 mole oxygen"/"32.0 g" = "1.904 "mu"moles oxygen"#

If you go by the number of moles of glucose, the reaction would have needed

#"0.3053 "mu"moles glucose" * "6 moles oxygen"/"1 mole glucose" = "1.832 "mu"moles oxygen"#

SInce you've got more oxygen than the reaction would have needed for that much glucose, glucose will act as a limiting reagent #-># only 1.832 micromoles of oxygen will take part in the reaction.

This means that you'll produce 1.832 micromoles of water, which is equivalent to a mass of

#"1.823 "mu"moles" * "18.02 g"/"1 mole water" = "32.9 "mu"g of water"#