Question #17e2e

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Question #17e2e

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Answer 1 · 2016-09-30T12:09:00

You are to determine the number of molecules of $N H_{3}$ $NH_3$ .

One molecule of $N H_{3}$ $NH_3$ contains 3 atoms of Hydrogen.So to produce one mole of $N H_{3}$ $NH_3$ we require $\frac{3}{2} m o l H_{2}$ $3/2molH_2$ The following equation also supports that

$N_{2} + 3 H_{2} \to 2 N H_{3}$ $N_2+3H_2->2NH_3$

So 1 mole $H_{2}$ $H_2$ produces $\frac{2}{3} m o l N H_{3}$ $2/3mol NH_3$
Hydrogen taken $= 7.98 \times 10^{- 4} g$ $=7.98xx10^-4g$

$= 7.98 \times 10^{- 4} g \times \frac{1}{2.016 \frac{g}{mol}}$ $=7.98xx10^-4gxx1/(2.016g/"mol")$

So no.of moles of $N H_{3}$ $NH_3$ produced
$= \frac{2}{3} \times 7.98 \times 10^{- 4} g \times \frac{1}{2.016 \frac{g}{mol}}$ $=2/3xx7.98xx10^-4gxx1/(2.016g/"mol")$

So no.of molecules of $N H_{3}$ $NH_3$ produced
$= \frac{2}{3} \times 7.98 \times 10^{- 4} g \times \frac{1}{2.016 \frac{g}{mol}} \times 6.023 \times 10^{23} \frac{molecule}{mol}$ $=2/3xx7.98xx10^-4gxx1/(2.016g/"mol")xx6.023xx10^23"molecule"/"mol"$
$= 15.9 \times 10^{19}$ $=15.9xx10^19$ molecules

Answer 2 · 2016-09-30T12:28:29

Here's what I got.

Explanation:

The problem wants you to find the number of molecules of ammonia, ${NH}_{3}$ $"NH"_3$ , not hydrogen gas, $H_{2}$ $"H"_2$ .

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

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

$= 2.38 * 10^20"molecules H"_2$

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

$color(blue)(3)"H"_ (2 (g)) + "N"_ (2(g)) -> color(purple)(2)"NH"_ (3(g))$

Notice that you have $color(blue)(3)$ molecules of hdyrogen gas reacting with $1$ molecule of nitrogengas to produce $color(purple)(2)$ 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 $color(blue)(3):color(purple)(2)$ mole ratio.

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

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

$= 1.59 * 10^(20)"molec NH"_3$

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.

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Question #17e2e

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