Question

Be sure to answer all parts. Ammonium nitrate (NH4NO3) 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 0.2101-g sample of industrially prepared NH4NO3 required 20.87 mL of 0.1073 M NaOH for neutralization. (a) Enter a net ionic equation for the reaction (include states of matter). NH, (aq) + OH (aq) --NH3(aq) +H,00) (b) What is the percent purity of the sample?

Answer 1

a.NH4NO3(aq) + NaOH(aq) ------------> NH3(g) + NaNO3(aq) + H2O(l)

NH4^+(aq) + NO3^-(aq) + Na^+(aq) + OH^-(aq) ------------> NH3(g) + Na^+(aq) + NO3^-(aq) + H2O(l)

removal of spectator ions to get net ionic equation

NH4^+(aq) + OH^-(aq) ------------> NH3(g)   + H2O(l)

b.

no of moles of NaOH = molarity * volume in L

                                  = 0.1073*0.02087 = 0.002239351moles

NH4NO3(aq) + NaOH(aq) ------------> NH3(g) + NaNO3(aq) + H2O(l)

1 mole of NaOH react with 1 mole of NH4NO3

0.00223935 moles of NaOH react with 0.00223935 moles of NH4NO3

mass of NH4NO3 = no of moles * gram molar mass

                           = 0.00223935*80.0432 = 0.179 g of NH4NO3

The percent purity of sample = 0.179*100/0.2101 = 85.2% >>>answer