# Question #49595

##### 1 Answer

#### 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 **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

#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 = 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 **total number of moles** of water vapor produced by the reactions is

#(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, **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 **mass** of ammonia and **mass** of hydrazine, you can say that equation

#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**.