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Data Table 3: 80-L Driver-Side Air Bag Activity Calculations Moles of CO 2 requi

ID: 549172 • Letter: D

Question

Data Table 3: 80-L Driver-Side Air Bag

Activity

Calculations

Moles of CO2 required to inflate 80-L driver-side air bag at room temperature and pressure

3.25moles

Balanced equation for the reaction of NaHCO3 and CH3COOH

NaHCO3+CHCOOH=CH3COONa+H2O+CO2

Grams of NaHCO3 needed for the reaction

273 g

mL of CH3COOH (0.833 M) needed for the reaction

3.9 L, 3900mL

Data Table 4: 160-L Front Passenger-Side Air Bag

Activity

Calculations

Moles of CO2 required to inflate 160-L front passenger-side air bag at room temperature and pressure

7.14 moles

Balanced equation for the reaction of NaHCO3 and CH3COOH

NaHCO3+CHCOOH=CH3COONa+H2O+CO2

Grams of NaHCO3 needed for the reaction

599.76, 600 grams

mL of CH3COOH (0.833 M) needed for the reaction

8.57L, 8470mL

Based upon the observed performance of the air bag models and the amounts of sodium bicarbonate and acetic acid (vinegar) needed for an automotive air bag of 80 or 160 L, are these reactants a good substitute for sodium azide? One additional note regarding sodium azide: the rate of inflation after a triggering impact is 40 milliseconds (0.04 s).

Activity

Calculations

Moles of CO2 required to inflate 80-L driver-side air bag at room temperature and pressure

3.25moles

Balanced equation for the reaction of NaHCO3 and CH3COOH

NaHCO3+CHCOOH=CH3COONa+H2O+CO2

Grams of NaHCO3 needed for the reaction

273 g

mL of CH3COOH (0.833 M) needed for the reaction

3.9 L, 3900mL

Explanation / Answer

Recall that

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

mol of N2 required to fill up 80 L --> 80/22.4 = 3.57 mol of N2

mol of NaN3 = 2/3*3.57 = 2.38 mol of NaN3

mass = mol*MW = 2.38*65.00987 = 154.72 g

Density of NaN3# --> 1.85 g/mL

V = masS/D = 154.72/1.85

V = 83.632 mL required

which is much less than 3.9 L and 8.57 L

that is...

this is still much more effective

avoid new "alternatives" remain with NaN3