# Question d111c

Apr 25, 2016

$1.20 \cdot {10}^{24} \text{molecules H"_2"O}$

#### Explanation:

Start by taking a look at the balanced chemical equation that describes this redox reaction

$\textcolor{red}{4} {\text{NH"_ (3(g)) + 3"O"_ (2(g)) -> 2"N"_ (2(g)) + color(blue)(6)"H"_ 2"O}}_{\left(l\right)}$

Since no mention of the number of molecules of oxygen gas available for this reaction was made in the problem, you can assume that oxygen gas is in excess.

This means that all the moles of ammonia, ${\text{NH}}_{3}$, will take part in the reaction.

As you can see, the reaction produces $\textcolor{b l u e}{6}$ moles of water for every $\textcolor{red}{4}$ moles of ammonia that take part in the reaction.

Now, a mole is simply a very large collection of molecules. More specifically, in order to have one mole of a substance, you need to have

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 mole" = 6.022 * 10^(23)"molecules}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ Avogadro's number

This means that you can convert the mole ratio to a molecule ratio by using Avogadro's number as a conversion factor. So, instead of $\textcolor{b l u e}{6}$ moles of water and $\textcolor{red}{4}$ moles of ammonia, you will have

$\textcolor{b l u e}{6} \textcolor{w h i t e}{a} \text{moles H"_2"O" = color(blue)(6) xx 6.022 * 10^(23)"molecules H"_2"O}$

$\textcolor{red}{4} \textcolor{w h i t e}{a} {\text{moles NH"_3 = color(red)(4) xx 6.022 * 10^(23)"molecules NH}}_{3}$

This will be equivalent to

$\left(\textcolor{b l u e}{6} \textcolor{w h i t e}{a} {\text{moles H"_2"O")/(color(red)(4)color(white)(a)"moles NH"_3) = (color(blue)(6) xx color(black)(cancel(color(black)(6.022 * 10^(23))))"molecules H"_2"O")/(color(red)(4) xx color(black)(cancel(color(black)(6.022 * 10^(23))))"molecules NH}}_{3}\right)$

As you can see, a mole ratio will always be equivalent to a molecule ratio because of the fact that a mole is simply a set number of molecules

color(purple)(|bar(ul(color(white)(a/a)color(black)((color(blue)(6)color(white)(a)"moles H"_2"O")/(color(red)(4)color(white)(a)"moles NH"_3) = (color(blue)(6)color(white)(a)"molecules H"_2"O")/(color(red)(4)color(white)(a)"molecules NH"_3)color(white)(a/a)|)))#

So, if the reaction uses up $8.02 \cdot {10}^{23}$ molecules of ammonia, it follows that you can expect it to produce

$8.02 \cdot {10}^{23} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{molec. NH"_3))) * (color(blue)(6)color(white)(a)"molec. H"_2"O")/(color(red)(4)color(red)(cancel(color(black)("molec. NH"_3)))) = color(green)(|bar(ul(color(white)(a/a)1.20 * 10^(24)"molec. H"_2"O} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.