Question

Question #49595

Answer 1

`"52% NH"_3`

Explanation:

!! LONG ANSWER !!

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

`4"NH"_ (3(g)) + 3"O"_ (2(g)) -> 2"N"_ (2(g)) + 6"H"_ 2"O"_ ((g))`

`"N"_ 2"H"_ (4(g)) + "O"_ (2(g)) -> "N"_ (2(g)) + 2"H"_ 2"O"_ ((g))`

Focus on the first reaction. Notice that you have

`"4 moles NH"_3 -> "2 moles N"_2 " and " 6 color(white)(a)"moles H"_2"O"`

This is equivalent to saying that

`"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

`xcolor(white)(a)"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

`"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

`ycolor(white)(a)"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 `(x/2 + y)`. The total number of moles of water vapor produced by the reactions is `((3x)/2 + 2y)`. This means that you have

`(x/2 + y)/((3x)/2 + 2y) = 2/5`

Rearrange to find

`(5x)/2 + 5y = 3x + 4y`

This is equivalent to

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

`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 `color(orange)("(*)")` 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 * m_x = 34.062 * m_y implies m_x = 34.062/32.0452 * m_y`

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

`m_"total" = m_x + m_y`

`m_"total" = 34.062/32.0452 * m_y + m_y = 66.1072/32.0452 * m_y`

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

`"% NH"_3 = ( (34.062 * color(red)(cancel(color(black)(m_y))))/color(red)(cancel(color(black)(32.0452)))) / ((66.1072 * color(red)(cancel(color(black)(m_y))))/color(red)(cancel(color(black)(32.0452)))) xx 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.