A 0.2947 g specimen containing sodium azide was treated with an unmeasured excess of sodium nitrite and 50.00?

Question

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 completetion of this reaction

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

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

Answer 1

Actually

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