Question #49595

!! 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.