For the reaction 2H_2 + O_2 -> 2H_2O, how many moles of water can be produced from 6.0 mol of oxygen?

Dec 5, 2015

$\text{12 moles H"_2"O}$

Explanation:

Your tools of choice for stoichiometry problems will always be the mole ratios that exist between the chemical species that take part in the reaction.

As you know, the stoichiometric coefficients attributed to each compound in the balanced chemical equation can be thought of as moles of reactants needed or moles of products formed in the reaction.

In your case, the balanced chemical equation for this synthesis reaction looks like this

2"H"_text(2(g]) + "O"_text(2(g]) -> color(red)(2)"H"_2"O"_text((l]])

Notice that the reaction requires $2$ moles of hydrogen gas and $1$ mole of oxygen gas to produce $\textcolor{red}{2}$ moles of water.

This tells you that the reaction produces twice as many moles of water as you have moles of oxygen gas that take part in the reaction.

You know that your reaction uses $6.0$ moles of oxygen. Assuming that hydrogen gas is not a limiting reagent, you can say that the reaction will produce

6.0 color(red)(cancel(color(black)("moles O"_2))) * (color(red)(2)" moles H"_2"O")/(1color(red)(cancel(color(black)("moles O"_2)))) = color(green)("12 moles H"_2"O")