# Question #a6390

Mar 23, 2015

The maximum mass of water that can be produced from your reaction is $\text{32.9." mu"g}$.

So, you're dealing with the combustion reaction of glucose, ${C}_{6} {H}_{12} {O}_{6}$; the balanced chemical equation for this reaction looks like this

${C}_{6} {H}_{12} {O}_{6} + \textcolor{red}{6} {O}_{2} \to 6 C {O}_{2} + \textcolor{red}{6} {H}_{2} O$

Notice the $1 : \textcolor{red}{6}$ 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

$\text{55.0"mu"g" * "1 mole glucose"/"180.16 g" = "0.3053 "mu"moles glucose}$

$\text{35.0"mu"g" * "1 mole oxygen"/"32.0 g" = "1.904 "mu"moles oxygen}$

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

$\text{0.3053 "mu"moles glucose" * "6 moles oxygen"/"1 mole glucose" = "1.832 "mu"moles oxygen}$

SInce you've got more oxygen than the reaction would have needed for that much glucose, glucose will act as a limiting reagent $\to$ 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

$\text{1.823 "mu"moles" * "18.02 g"/"1 mole water" = "32.9 "mu"g of water}$