Titration question.?

Question

Titration question regarding calculating percent mass created?

A 0.2947 g specimen containing sodium azide was treated with an unmeasured excess of sodium nitrite (NaN3) and 50.00 mL of 0.1030 M HClO4. Upon completion of this reaction

2H+ + N3- +NO_2- -----> H2O + N2O (gas) + N2 (gas)

the excess acid was titrated with 13.49 mL of 0.0800 M NaOH. Calculate the percentage of NaN3 in the sample.

Answer 1

find moles of standard:

0.050 litres of 0.1031 mol/Litre H+ = 0.005155 moles H+

find moles of NaOH back titrated:

0.01349 Litres @ 0.0800mol/litre = 0.001079 moles of NaOH

subtract the NaOH moles, from the H+ moles to find the moles of H+ that reacted with the azide:

0.005155 - 0.001079 = 0.004076 moles of H+ reacted with the azide

since: 2H+ + N3- +NO_2- -----> H2O + N2O (gas) + N2,...

0.004076 mol H+ @ 1 mole N3 / 2 mole H+ = 0.002038 moles of azide must have been present

use molar mass to get grams of NaN3:

0.002038 mol NaN3 @ 65.01 g/mol = 0.1325 grams of NaN#

Calculate the percentage of NaN3 in the sample:

0.1325 g NaN3 / 0.2947 g sample times 100 = 44.96%

your answer: 44.96 % NaN3