Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

1) Automobile airbags are inflated upon a serious impact. The impact triggers th

ID: 855956 • Letter: 1

Question

1) Automobile airbags are inflated upon a serious impact. The impact triggers the following chemical reaction: 2 NaN3(s) ¾? 2 Na(s) + 3 N2(g) If an automobile airbag has a total volume of 1.50 L, then (a) How many moles of N, are needed to inflate the airbag to 1.10 atm at 22.0 degree C? (b) What mass of NaN3 is required to inflate the airbag? (Hint. Step 1: USC the P. V.T data and the ideal gas law to find moles N2. Step 2: Use the reaction stoichiometry to convert moles N2 into moles NaN3. Step 3: use the molar mass of NaN3 and the moles of NaN3 to find the grams of NaN3). 2) Consider the reaction: P4(s) + 5 O2(g) ¾? P4O10(s) how many liters of oxygen gas at P = 730 torr and T = 18.0 degree C are required to form 94.6 g of P4O10? (Hint: Step 1: Find the number of moles in 94.6 g P4O10. Step2: use reaction stoichiometry to convert the moles of P4O10 into moles of O2. Siep3: Use ideal gas law and pressure and temperature data to solve for the volume of O2. The units of P and T must be converted into the same units as R.)

Explanation / Answer

1) PV= nRT

n = RT/PV

n = (0.0821*295/1.10*1.50)

n= 14.67 moles

n= mass/molar mas

14.67= mass/(42)

mass= 616.49 gr

2)

no.of moles of P4O10 = 94.6/304 = 0.311MOLES

1 MOLE OF P4 REACT WITH------------------------------------------5 MOLES OF OXYGEN

0.311 MOLES OF P4 REACT WITH---------------------------------------------?

= 0.311*5/1= 1.555 MOLES OF OXYGEN

V= nRT/P

V= 1.55*0.0821*291/(730/760)

V= 38.55 L OF OXYGEN