•If 65.1 L at STP of N2 gas are needed to inflate a real air bag to the proper size, how many grams of NaN3 must be included in the real air bag to generate this amount of N2? How many molecules of sodium are produced?

•Determine the percent composition of nitrogen in 26g of NaN3. Show your work.

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ayoushivt ayoushivt · Answer 1 · 2022-06-20T10:55:34+02:00

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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