Question

Ammonium nitrate (NH4N03) is one of the most important nitrogen-containing fertilizers. Its purity can be analyzed by titrating a solution of NH4NO3 with a standard NaOH solution. In one experiment a 0.2111-g sample of industrially prepared NH4NO3 required 22.62 mL of 0.1023 M NaOH for neutralization (a) Enter a net ionic equation for the reaction (include states of matter) (b) What is the percent purity of the sample?

Answer 1

Answer: The percentage purity of sample is 87.54 %.

Step-by-step explanation:

• For a:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of ammonium nitrate and sodium hydroxide is given as:

`NH_4NO_3(aq.)+NaOH(aq.)\rightarrow NH_3(g)+H_2O(l)+NaNO_3(aq.)`

Ionic form of the above equation follows:

`NH_4^+(aq.)+NO_3^-(aq.)+Na^+(aq.)+OH^-(aq.)\rightarrow NH_3(g)+H_2O(l)+Na^+(aq.)+NO_3^-(aq.)`

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

`NH_4^+(aq.)+OH^-(aq.)\rightarrow NH_3(g)+H_2O(l)`

• For b:

To calculate the moles of sodium hydroxide, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}`

Molarity of sodium hydroxide = 0.1023 M

Volume of sodium hydroxide = 22.62 mL = 0.02262 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

`0.1023mol/L=\frac{\text{Moles of sodium hydroxide}}{0.02262L}\\\\\text{Moles of sodium hydroxide}=0.00231mol`

For the given chemical equation:

`NH_4NO_3(aq.)+NaOH(aq.)\rightarrow NH_3(g)+H_2O(l)+NaNO_3(aq.)`

By Stoichiometry of the reaction:

1 mole of sodium hydroxide reacts with 1 mole of ammonium nitrate.

So, 0.00231 moles of sodium hydroxide will react with =
`(1)/(1)* 0.00231=0.00231mol` of ammonium nitrate.

• To calculate the mass of ammonium nitrate, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Moles of ammonium nitrate = 0.00231 mol

Molar mass of ammonium nitrate = 80 g/mol

Putting values in above equation, we get:

`0.00231mol=\frac{\text{Mass of ammonium nitrate}}{80g/mol}\\\\\text{Mass of ammonium nitrate}=0.1848g`

• To calculate the percentage purity of ammonium nitrate, we use the equation:

`\%\text{ purity}=\frac{\text{Mass of pure sample}}{\text{Mass of given sample}}* 100`

Mass of pure sample = 0.1848 g

Mass of given sample = 0.2111 g

Putting values in above equation, we get:

`\%\text{ purity of sample}=(0.1848g)/(0.2111g)* 100\\\\\%\text{ purity of sample}=87.54\%`

Hence, the percentage purity of sample is 87.54 %.