In the formation of smog, nitrogen and oxygen gas react to form nitrogen dioxide: N2(g)+2O2(g)→2NO2(g)
Part A How many grams of NO2 will be produced when 2.0 L of nitrogen at 870 mmHg and 29 ∘C are completely reacted? Express your answer using two significant figures. mNO2 m N O 2 =
When sensors in a car detect a collision, they cause the
reaction of sodium azide, NaN3, which generates nitrogen gas to
fill the air bags within 0.03 s. |
Part A How many liters of N2 are produced at STP if the air bag contains 137 g of NaN3? Express your answer with the appropriate units.
|
1)Solution- Using equation
n = PV / RT
= (870 mmHg) x (2.0 L) / ((62.36367 L Torr/K mol) x (29 + 273)K)
= 0.0924 mol N2
(0.0924 mol N2) x (2 mol NO2 / 1 mol N2) x (46.00550 g NO2/mol)
=8.502g NO2
2)
Molar mass of NaN3 = 65.0099 gm/mole
Now 137 gm of NaN3 = 137 / 65.0099 = 2.10 mole
From the reaction 2 mole of NaN3 produce 3 moles of N2 then 2.10
mole of NaN3 produce 2.10 X 3 / 2 = 3.15 mole of N2
At STP 1 mole gas occupy volume
= 22.414 liter
So 3.15 mole of N2 gas occupy volume = 3.15 X 22.414
= 70.60 liter
In the formation of smog, nitrogen and oxygen gas react to form nitrogen dioxide: N2(g)+2O2(g)→2NO2(g) Part...
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