# Question e5b9f

Mar 9, 2017

$\text{24 moles H"_2"O}$

#### Explanation:

The idea here is that the stoichiometric coefficients present in the balanced chemical equation can be used as mole ratios.

$\textcolor{b l u e}{4} {\text{NH"_ (3(g)) + color(darkorange)(7)"O"_ (2(g)) -> 4"NO"_ (2(g)) + color(purple)(6)"H"_ 2"O}}_{\left(l\right)}$

Notice that the reaction consumes $\textcolor{b l u e}{4}$ moles of ammonia and $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{7}$ moles of oxygen gas in order to produce $\textcolor{p u r p \le}{6}$ moles of water.

The first thing you need to do here is to figure out if one of the two reactants acts as a limiting reagent, i.e. if it gets consumed before all the moles of the second reactant get the chance to react.

In this case, $16.00$ moles of ammonia would require

16.00 color(red)(cancel(color(black)("moles NH"_3))) * (color(darkorange)(7)color(white)(.)"moles O"_2)/(color(blue)(4)color(red)(cancel(color(black)("moles NH"_3)))) = "28 moles O"_2#

which is exactly the amount of oxygen gas you have available. This tells you that both reactants will be completely consumed by the reaction.

Pick one of the two reactants and use it to determine the number of moles of water produced by the reaction

$16.00 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NH"_3))) * (color(purple)(6)color(white)(.)"moles H"_2"O")/(color(blue)(4)color(red)(cancel(color(black)("moles NH"_3)))) = color(darkgreen)(ul(color(black)("24 moles H"_2"O}}}}$

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