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.

Idont know. if you want an answer you need to give us some context dude.

Answer is: hydrogen is being oxidized.chemical reaction: co₂ + h₂ → co + h₂o.oxidation half reaction: h₂ → 2h⁺ + 2e⁻.reduction half reaction: c⁺⁴ + 2e⁻ → c⁺².oxidation is increase of oxidation number, hydrogen change oxidation number from neutral charge (in molecule of hydrogen) to oxidation number +1 (in water).