# Question 49595

Aug 7, 2016

${\text{52% NH}}_{3}$

#### Explanation:

Start by writing the balanced chemical equations that describe these combustion reactions

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

${\text{N"_ 2"H"_ (4(g)) + "O"_ (2(g)) -> "N"_ (2(g)) + 2"H"_ 2"O}}_{\left(g\right)}$

Focus on the first reaction. Notice that you have

$\text{4 moles NH"_3 -> "2 moles N"_2 " and " 6 color(white)(a)"moles H"_2"O}$

This is equivalent to saying that

$\text{1 mole NH"_3 -> 1/2 color(white)(a)"moles N"_2 " and "3/2color(white)(a) "moles H"_2"O}$

Let's assume that $x$ represents the number of moles of ammonia present in the original mixture. You can say that you have

$x \textcolor{w h i t e}{a} \text{moles NH"_3 -> (x/2) color(white)(a)"moles N"_2 " and "((3x)/2)color(white)(a) "moles H"_2"O}$

Now focus on the second reaction. You have

$\text{1 mole N"_2"H"_4 -> "1 mole N"_2 " and " 2color(white)(a)"moles H"_2"O}$

If you take $y$ to be the number of moles of hydrazine present in the original mixture, you can say that you have

$y \textcolor{w h i t e}{a} \text{moles N"_2"H"_4 -> ycolor(white)(a)"moles N"_2 " and " (2y)color(white)(a)"moles H"_2"O}$

Since all gases are kept under the same conditions for pressure and temperature, you know that the volume ratio that exists between the products is equivalent to a mole ratio, as described by Avogadro's Law.

In your case, a $4 : 10$ volume ratio between the volume of nitrogen gas and the volume of hydrogen gas produced by the reactions tells you that you have

4:10 = 2/5 = "total moles of N"_2/("total moles of H"_2"O")

The total number of moles of nitrogen gas produced by the reactions is $\left(\frac{x}{2} + y\right)$. The total number of moles of water vapor produced by the reactions is $\left(\frac{3 x}{2} + 2 y\right)$. This means that you have

$\frac{\frac{x}{2} + y}{\frac{3 x}{2} + 2 y} = \frac{2}{5}$

Rearrange to find

$\frac{5 x}{2} + 5 y = 3 x + 4 y$

This is equivalent to

$5 x + 10 y = 6 x + 8 y$

x = 2y" " " "color(orange)("(*)")

You now know that the original mixture contained twice as many moles of ammonia, $x$, than moles of hydrazine, $y$. To find the percent composition of ammonia in the original mixture, use the molar masses of the two compounds.

M_("M NH"_3) = "17.031 g mol"^(-1)

M_("M N"_2"H"_4) = "32.0452 g mol"^(-1)

If you take ${m}_{x}$ to be the mass of ammonia and ${m}_{y}$ to be the mass of hydrazine, you can say that equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$ gives you

overbrace((m_x color(red)(cancel(color(black)("g"))))/(17.031 color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)("mol"^(-1))))))^(color(blue)("no. of moles of NH"_3)) = overbrace(2 * (m_y color(red)(cancel(color(black)("g"))))/(32.0452 color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)("mol"^(-1))))))^(color(purple)("moles of N"_2"H"_4))

Therefore, you have

$32.0452 \cdot {m}_{x} = 34.062 \cdot {m}_{y} \implies {m}_{x} = \frac{34.062}{32.0452} \cdot {m}_{y}$

You know that the total mass of the mixture must have been

${m}_{\text{total}} = {m}_{x} + {m}_{y}$

${m}_{\text{total}} = \frac{34.062}{32.0452} \cdot {m}_{y} + {m}_{y} = \frac{66.1072}{32.0452} \cdot {m}_{y}$

Therefore, the percent composition of ammonia in the original mixture was

${\text{% NH}}_{3} = \frac{\frac{34.062 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{m}_{y}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{32.0452}}}}}{\frac{66.1072 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{m}_{y}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{32.0452}}}}} \times 100$

"% NH"_3 = 34.062/66.1072 * 100 = color(green)(|bar(ul(color(white)(a/a)color(black)(52%)color(white)(a/a)|)))#

The answer is rounded to two sig figs.