Question

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. 2NaN3(s)→2Na(s)+3N2(g)

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. V V =

Answer 1

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