How many grams of water could be made from 5.0 mol #H_2# and 3.0 mol #O_2#?

1 Answer
Sep 2, 2017

Answer:

Approx. #90*g# of water will be evolved.

Explanation:

We need a stoichiometric equation as our starting point.....

#H_2 +1/2O_2 rarr H_2O#

The stoichiometric equation unequivocally tells me that #2 *g# of dihydrogen reacts with #16 *g# dioxygen to give #18*g# of water.

We have a #3.0*mol# quantity of #O_2#, and a #5.0*mol# quantity of #H_2#. Clearly, #H_2# is THE LIMITING REAGENT, the reagent in stoichiometric deficiency. Do you agree?

And so ONLY #2.5*mol# of #O_2# will react to give (again I use the stoichiometry of the given reaction) #5*mol# of water......

i.e. #5*cancel(mol)xx18.01*g*cancel(mol^-1)=90*g#

This is something worth spending time on if you don't see it immediately. This is an example of stoichiometric equivalence, where a MASS of reagent, dihydrogen, dioxygen, etc. REPRESENTS a NUMBER of dihydrogen, dioxygen molecules, which is what a molar quantity is. If you have another query.......ask.