The percent composition of nitrogen in 26g of NaN₃ is 21.5 % , 126.1 gm of NaN₃ is required and 1.94 moles of Na is produced.
Mole is a measurement unit which relate the molar mass with the number of molecules /atoms.
It is given that
If 65.1 L at STP of N₂ gas are needed to inflate a real air bag to the proper size
1 * 65.1 = n * 0.082 * 273
n = 2.91 moles
2NaN₃ --->2Na +3 N₂
Mole fraction ratio of nitrogen to NaN₃ is 3 : 2
Therefore to produce 2.91 moles of nitrogen gas ,
2.91 / x = 3/2
2.91 * 2/3 = x
x = 1.94 moles
1.94 moles of Sodium is produced
1.94* 65 = 126.1 gm of NaN₃
the percent composition of nitrogen in 26g of NaN₃
65 grams means 1 mole of NaN₃
26 grams will mean 26/65 = 0.4 moles
Mass percentage of nitrogen = Molar mass of Nitrogen/ Molar mass of NaN₃
Molar mass of Nitrogen = 14 gmol-1
Molar mass of NaN₃ = 23 + ( 3 x 14 ) = 65 gmol-1
Mass percentage of nitrogen = 14 gmol-1/ 65 g mol-1
= 21.5 %
In 26 grams , 5.6 grams of Nitrogen is present.
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