Taj asked in Science & MathematicsChemistry · 1 decade ago

What mass of NaN3(s) must be reacted in order to inflate an air bag to 70.5 L at STP?

Air bags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. This causes sodium azide (NaN3) to decompose explosively according to the following reaction.

2 NaN3(s) 2 Na(s) + 3 N2(g)

What mass of NaN3(s) must be reacted in order to inflate an air bag to 70.5 L at STP?

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  • Favorite Answer

    At STP, one mole of any gas occupies a space of 22.4 L:

    70.5 L * 1 mole N2/22.4 L = 3.15 moles of N2

    Now we know how many moles of N2 gas we need from the reaction. Multiply this by the mole ration from the reaction equation:

    3.15 mol N2 * (2 mol NaN3)/(3 mol N2) = 2.10 mol NaN3

    This tells us that 2.10 moles of NaN3 need to be reacted to form the 3.15 moles (70.5 L) of N2 gas that we need to inflate the airbag to 70.5 L.

    2.10 mol NaN3 * 65.01 gNaN3/mol NaN3 = 137 g NaN3

    *** 137 grams of NaN3 need to be reacted to inflate the airbag to 70.5 L at STP.

    Source(s): 10th grade chemistry, college chemistry. I'm in the second semester of organic chemistry.
  • 4 years ago

    R= .0821 V = 78.7 L T= 273 P = 1 atm find #moles of N2 gas by using gas equation PV=nRt n = RT/PV substitute above values, n = .0821 x 273 / 78.7 = 0.2648 moles of N2 gas Note the stoichiometry from the balanced equation (2 moles of NaN3 react to produce 3 moles of N2 gas), and calculate # of moles of NaN3 that need to explode 2NaN3 --> 2Na + 3N2 0.2648 moles N2 x 2NaN3/3N2 = 0.1765 moles NaN3 Find MW of NaN3, MW NaN3 = 65 65 g/mole x 0.1765 moles = 11.47 g NaN3 answer

  • 1 decade ago

    2NaN3 --> 2Na + 3N2

    70.5 L at STP = 70.5/22.4moles of N2 or 3.147 moles of N2

    You get 1.5 mole of N2 from 1 mole of NaN3, so you require 3.147/1.5 = 2.098 moles of NaN3.

    1 mole of NaN3 = 65g, so you need-

    2.1moles x 65g/mole = 136.5 g of NaN3

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