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

Sep 2, 2017

#### Answer:

Approx. $90 \cdot g$ of water will be evolved.

#### Explanation:

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

${H}_{2} + \frac{1}{2} {O}_{2} \rightarrow {H}_{2} O$

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

We have a $3.0 \cdot m o l$ quantity of ${O}_{2}$, and a $5.0 \cdot m o l$ quantity of ${H}_{2}$. Clearly, ${H}_{2}$ is THE LIMITING REAGENT, the reagent in stoichiometric deficiency. Do you agree?

And so ONLY $2.5 \cdot m o l$ of ${O}_{2}$ will react to give (again I use the stoichiometry of the given reaction) $5 \cdot m o l$ of water......

i.e. $5 \cdot \cancel{m o l} \times 18.01 \cdot g \cdot \cancel{m o {l}^{-} 1} = 90 \cdot g$

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.