Question

A lab chemist studying the reaction rate of the decomposition of
sodium azide in nitrogen gas requires a 400 mL stock solution of 0.45 M
sodium azide.
a.) How many grams of sodium azide are required to prepare this
solution?
b.) If the reaction experiment requires 125 mL of sodium azide and
88% of the sodium azide decomposes during the process of the
reaction, what is the final concentration of sodium azide assuming the volume is unchanged?
the volume is unchanged?

Answer 1

Step 1:

Calculation of moles of sodium azide

Mol = V (L) x M

n NaN3 = 0.400 L x 0.45 M

            =0.18 mol

Calculation of mass of NaN3

m NaN3

m = mol x m M

m NaN3

            =0.18 mol x 65.009 g mol^-1

            = 17.70 g

Mass of sodium azide is 17.70 g

b) Final concentration = [ Initial n NaN3 ( 400 mL ) – decomposed n NaN3] / V of soln in L

        Calculation of final moles

Moles dissociated = (0.145 L x 0.450 M ) x 88 / 100 = 0.1584 mol = 0.05742 mol

Final concentration = (0.05742 mol / 0.145 L ) = 0.396 = 0.40 M