Reacting Ammonia with Oxygen

## **Problem**

Consider the chemical reaction

4

NH

3

5

O

2

⟶
6

H

2

O
4

NO
905

kJ

{\displaystyle \ce{4 NH3 + 5 O2 -> 6 H2O + 4 NO + 905 kJ }}

**Part 1:** What mass of

NO

{\displaystyle \ce{NO}}

 (nitrogen monoxide) is produced when

2.0

m
o
l

{\displaystyle 2.0 \; mol}

 of

NH

3

{\displaystyle \ce{NH3}}

 (ammonia) react?

**Part 2:** What volume of

NH

3

{\displaystyle \ce{NH3}}

 is required to react with

3.00

L

{\displaystyle 3.00 \; L}

 of oxygen?

**Part 3:** What volume of

H

2

O

{\displaystyle \ce{H2O}}

 is produced along with

2.84

L

{\displaystyle 2.84 \; L}

 of

NO

{\displaystyle \ce{NO}}

?

**Part 4:** How much energy is produced when

2.70

g

{\displaystyle 2.70 \; g}

 of

NH

3

{\displaystyle \ce{NH3}}

 are burned?

**Part 5:** If

200

k
J

{\displaystyle 200 \; kJ }

 of heat is produced, what mass of both products are also formed?

## **Solution**

**Part 1**

Start by calculating the molar masses of

NO

{\displaystyle \ce{NO}}

 and

NH

3

{\displaystyle \ce{NH3}}

,

 Oxygen

Add the net molar mass of each element to obtain the net molar mass of nitrogen monoxide.

14.01

g

m
o
l

16.00

g

m
o
l

=
30.01

g

m
o
l

{\displaystyle 14.01 \; \frac{g}{mol} + 16.00 \; \frac{g}{mol} = 30.01 \; \frac{g}{mol} }

 Hydrogen

Add the net molar mass of each element to obtain the net molar mass of ammonia.

14.01

g

m
o
l

3.03

g

m
o
l

=
17.04

g

m
o
l

{\displaystyle 14.01 \; \frac{g}{mol} + 3.03 \; \frac{g}{mol} = 17.04 \; \frac{g}{mol} }

Since the coefficients of

NO

{\displaystyle \ce{NO}}

 and

NH

3

{\displaystyle \ce{NH3}}

 are both 4, we get

2.0

m
o
l

N

H
3

×

4

m
o
l

N
O

4

m
o
l

N

H
3

×
30.01

g

m
o
l

=
60.02

g

N
O

{\displaystyle 2.0 \; mol\; \mathrm{NH\_3} \times \frac{4 \; mol\; \mathrm{NO}}{4 \; mol\; \mathrm{NH\_3}} \times 30.01 \; \frac{g}{mol} = 60.02 \; g \; \mathrm{NO} }

Hence, the mass of

NO

{\displaystyle \ce{NO}}

 produced by reacting 2 moles of

NH

3

{\displaystyle \ce{NH3}}

 is roughly 60 grams.

**Part 2**

The coefficients of

NH

3

{\displaystyle \ce{NH3}}

 and

O

2

{\displaystyle \ce{O2 }}

 are 4 and 5 respectively, so

4

m
o
l

N

H
3

5

m
o
l

O
2

×
3.00

L

O
2

=
2.40

L

N

H
3

{\displaystyle \frac{4 \; mol\; \mathrm{NH\_3}}{5 \; mol\; \mathrm{O\_2}} \times 3.00 \; L \; \mathrm{O\_2} = 2.40 \; L \; \mathrm{NH\_3} }

**Part 3**

The coefficients of

NO

{\displaystyle \ce{NO}}

 and

H

2

O

{\displaystyle \ce{H\_2O }}

 are 4 and 6 respectively, so

6

m
o
l

H
2

O

4

m
o
l

N
O

×
2.84

L

N
O

=
4.26

L

H
2

O

{\displaystyle \frac{6 \; mol\; \mathrm{H\_2O}}{4 \; mol\; \mathrm{NO}} \times 2.84 \; L \; \mathrm{NO} = 4.26 \; L \; \mathrm{H\_2O} }

**Part 4**

Determine the number of moles for

2.70

g

{\displaystyle 2.70 \; g}

 of

NH

3

{\displaystyle \ce{NH3}}

2.7

g

17.04

g

/

m
o
l

=
0.1585

m
o
l

{\displaystyle \frac{2.7 \; g}{17.04 \; g/mol} = 0.1585 \; mol }

Comparing this with the 4 moles of

N

H
3

{\displaystyle NH\_3}

, we find

x

905

k
J

=

0.1585

m
o
l

N

H
3

4

m
o
l

N

H
3

{\displaystyle \frac{x }{905 \; kJ} = \frac{0.1585 \; mol \; \mathrm{NH\_3}}{4 \; mol \; \mathrm{NH\_3}} }

Solving this equation yields

# x
35.8

k
J

{\displaystyle x = 35.8 \; kJ }

.

**Part 5**

First determine the ratio of

200

k
J

{\displaystyle 200 \; kJ }

 to

905

k
J

{\displaystyle 905 \; kJ }

.

200

k
J

905

k
J

=
0.2210

{\displaystyle \frac{200 \; kJ}{905 \; kJ} = 0.2210 }

Next calculate the number of moles for

NO

{\displaystyle \ce{NO}}

and

H

2

O

{\displaystyle \ce{H2O}}

. The coefficients of

NO

{\displaystyle \ce{NO}}

 and

H

2

O

{\displaystyle \ce{H\_2O }}

 are 4 and 6 respectively, so

0.2210
×
4

m
o
l

N
O

=
0.8840

m
o
l

N
O

0.2210
×
6

m
o
l

H
2

O

=
1.3260

m
o
l

H
2

O

{\displaystyle \begin{align}
&0.2210 \times 4 \;mol\; \mathrm{NO} = 0.8840 \; mol \; \mathrm{NO} \\[5 pt]
&0.2210 \times 6 \;mol\; \mathrm{H\_2O} = 1.3260 \; mol \; \mathrm{H\_2O}
\end{align} }

Then calculate the molar mass of

H

2

O

{\displaystyle \ce{H2O}}

 (the molar mass of

NO

{\displaystyle \ce{NO}}

 was calculated in part 1).

 Oxygen

Add the net molar mass of each element to obtain the net molar mass of water.

2.02

g

m
o
l

16.00

g

m
o
l

=
18.02

g

m
o
l

{\displaystyle 2.02 \; \frac{g}{mol} + 16.00 \; \frac{g}{mol} = 18.02 \; \frac{g}{mol} }

Lastly, colculate the mass of each molecule.

30.01

g

m
o
l

×
0.8840

m
o
l

N
O

=
26.52

g

N
O

18.03

g

m
o
l

×
1.3260

m
o
l

H
2

O

=
23.89

g

H
2

O

{\displaystyle \begin{align}
&30.01 \; \frac{g}{mol} \times 0.8840 \;mol\; \mathrm{NO} = 26.52 \; g \; \mathrm{NO} \\[5 pt]
&18.03 \; \frac{g}{mol} \times 1.3260 \;mol\; \mathrm{H\_2O} =23.89 \; g \; \mathrm{H\_2O}
\end{align} }

Therefore, 26.5 grams of

NO

{\displaystyle \ce{NO}}

 and 23. 9 grams of

H

2

O

{\displaystyle \ce{H2O}}

 are formed.