Automobile air bags inflate following a serious impact. The impact triggers the following chemical reaction. 2NaN3(s)->2Na(s)+3N2(g)

If an automobile air bag has a volume of 11.9 L, what mass of NaN3 (in g) is required to fully inflate the air bag upon impact? Assume STP conditions.

Explanation:

To determine the molar volume of the gas according to the equation at stp;

1 mole = 22.4 dm³

2 moles = 44.8 dm³

To determine the mass of NaN₃ inflated according to the equation

2NaN₃ (where Na = 23g and N = 14) = (2 × 23) + 2(14 × 3)

= 130 g

Hence, if 130g of NaN₃ is required to inflate 44.8 dm³ airbag upon impact

what mass of NaN₃ is required to fully inflate the air bag upon impact;

130g ⇒ 44.8 dm³

? ⇒ 11.9 dm³ (dm³ is same as L)

? = 130 × 11.9/44.8

? = 34.5g

34.5g of NaN₃ is required to fully inflate 11.9 L of air bag upon impact

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.