# Question 17e2e

Sep 30, 2016

You are to determine the number of molecules of $N {H}_{3}$ .

One molecule of $N {H}_{3}$ contains 3 atoms of Hydrogen.So to produce one mole of $N {H}_{3}$ we require $\frac{3}{2} m o l {H}_{2}$ The following equation also supports that

${N}_{2} + 3 {H}_{2} \to 2 N {H}_{3}$

So 1 mole ${H}_{2}$ produces $\frac{2}{3} m o l N {H}_{3}$
Hydrogen taken $= 7.98 \times {10}^{-} 4 g$

$= 7.98 \times {10}^{-} 4 g \times \frac{1}{2.016 \frac{g}{\text{mol}}}$

So no.of moles of $N {H}_{3}$ produced
$= \frac{2}{3} \times 7.98 \times {10}^{-} 4 g \times \frac{1}{2.016 \frac{g}{\text{mol}}}$

So no.of molecules of $N {H}_{3}$ produced
=2/3xx7.98xx10^-4gxx1/(2.016g/"mol")xx6.023xx10^23"molecule"/"mol"
$= 15.9 \times {10}^{19}$molecules

Sep 30, 2016

Here's what I got.

#### Explanation:

The problem wants you to find the number of molecules of ammonia, ${\text{NH}}_{3}$, not hydrogen gas, ${\text{H}}_{2}$.

Your calculation is set up to find the number of molecules of hydrogen gas present in the $7.98 \cdot {10}^{- 4} \text{g}$ sample.

7.98 * 10^(-4)cancel("g H"_2) * (1cancel("mole H"_2))/(2.016cancel("g H"_2)) * (6.022 *10^23"molec H"_2)/(1cancel("mole H"_2))

$= 2.38 \cdot {10}^{20} {\text{molecules H}}_{2}$

As you can see, you're missing a conversion factor to take you from molecules of hydrogen gas to molecules of ammonia.

This conversion factor comes from the balanced chemical equation that describes the synthesis of ammonia

$\textcolor{b l u e}{3} {\text{H"_ (2 (g)) + "N"_ (2(g)) -> color(purple)(2)"NH}}_{3 \left(g\right)}$

Notice that you have $\textcolor{b l u e}{3}$ molecules of hdyrogen gas reacting with $1$ molecule of nitrogengas to produce $\textcolor{p u r p \le}{2}$ molecules of ammonia.

You can thus say that if you start with $x$ molecules of hydrogen gas, you can get to molecules of ammonia by using

xcolor(white)(a)cancel("molecules H"_2) * (color(purple)(2)color(white)(a)"molecules NH"_3)/(color(blue)(3)cancel("molecules H"_2)) = (2/3x)" molecules NH"_3

In your case, you will have

2.38 * 10^(20)cancel("molec H"_2) * (color(purple)(2)color(white)(a)"molec NH"_3)/(color(blue)(3)cancel("molec H"_2)) = 1.59 * 10^(20)"molec NH"_3

In fact, you don't need to find the number of molecules of hydrogen gas, all you need to do is use the aforementioned $\textcolor{b l u e}{3} : \textcolor{p u r p \le}{2}$ mole ratio.

Since a mole is simply a fixed number of molecules, you can say that

7.98 * 10^(-4)cancel("g H"_2) * (1cancel("mole H"_2))/(2.016cancel("g H"_2)) * (color(purple)(2)cancel("moles NH"_3))/(color(blue)(3)cancel("moles H"_2)) * (6.022 *10^23"molec NH"_3)/(1cancel("mole NH"_3))#

$= 1.59 \cdot {10}^{20} {\text{molec NH}}_{3}$

So remember, don't jump in head-first, make sure you understand what the problem wants you to find first.

Once you have a clear understanding of what you have and waht you need to calculate, organize your conversion factors and plug in your values.