If 65.1l of n2(g) is needed to properly inflate an airbag, how many grams of nan3(s) are needed for the reaction

The equation for this reaction is as follow,

                                2 NaN₂ (s)    →    2 Na (s) + 3 N₂ (g)

According to equation,

           67.2 L N₂ (3 moles) is produced by  =  102 g (2 moles) NaN₂
So,
                    65.1 L N₂ will be produced by  =  X g of NaN₂

Solving for X,
                      X  =  (102 g × 65.1 L) ÷ 67.2

                      X  =  98.81 g of NaN₂

Result:
           98.81 g of NaN₂ is required to produce 65.1 L N₂ to inflate airbag.

It is 4.92 e-4 sorry if i am wrong