Question

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

Answer 1

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.