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USPshnik [31]
3 years ago
11

If 65.1L of N2(g) is needed to properly inflate an airbag, how many grams of NaN3(s) are needed for the reaction

Chemistry
1 answer:
Vladimir [108]3 years ago
3 0
The equation for this reaction is as follow,

<span>                                2 </span>NaN₂ (s)    →    2 Na (s<span>) + 3 </span>N₂ (g<span>)
</span>
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.
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