Question 45a2e

Jun 28, 2017

${\text{3.1 moles O}}_{2}$

Explanation:

The thing to remember about a balanced chemical equation is that the stoichiometric coefficients that are added to each chemical species can be treated as moles.

In your case, you know that

$\textcolor{b l u e}{2} {\text{H"_ 2"O"_ ((l)) -> 2"H"_ (2(g)) + "O}}_{2 \left(g\right)}$

Notice that it takes $\textcolor{b l u e}{2}$ molecules of water to produce $2$ molecules of hydrogen gas and $1$ molecule of oxygen gas. Since a mole is simply a very large collection of molecules, you can say that it takes $\textcolor{b l u e}{2}$ moles of water to produce $2$ moles of hydrogen gas and $1$ mole of oxygen gas.

This is equivalent to saying that water and oxygen gas have a $\textcolor{b l u e}{2} : 1$ mole ratio in this reaction, i.e. at 100%# yield, the reaction will produce half as many moles of oxygen gas as the number of moles of water that take part in the reaction.

This means that $6.2$ moles of water will produce

$6.2 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles H"_2"O"))) * "1 mole O"_2/(color(blue)(2)color(red)(cancel(color(black)("moles H"_2"O")))) = color(darkgreen)(ul(color(black)("3.1 moles O}}_{2}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the number of moles of water.

So remember, the stoichiometric coefficients can be referred to as moles. The chemical formulas represent the reactants and the products that take part in the reaction.