6-2 Engineering a Better Air Bag Lab Report

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Southern New Hampshire University *

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Chemistry

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Oct 30, 2023

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Engineering a Better Airbag Ashley Shaw 06/11/23 Data Use 0.0821 L atm mol K for R. Data Table 1: Model Air Bag Activity Data and Calculations Volume of 6 × 9 inch bag 1.20 L Room temperature (in K) 22C + 273.15 = 295.15K Room pressure (in atm) 29.77 inHg 29.921 atm / inHg = 0.9949 atm Moles of CO 2 required to inflate bag at room temperature and pressure. Show work, including all units, in the space provided. 0.9949 atmx 1.20 L 0.0821 atm L mol K x 295.15 K = 1.19388 atm L 24.231815 atm L mol = 0.04926 mol CO 2 Balanced equation for the reaction of NaHCO 3 and CH 3 COOH NaHCO 3 (s)+ CH 3 COOH(aq) H 2 O(l) + CO 2 (g) + C 2 H 3 NaO 2 (aq) Na 1:1 H 5:5 C 3:3 O 5:5 Mass of NaHCO 3 needed for the reaction. Show work, including all units, in the space provided. Grams = 0.04926 mol CO 2 x 1mol NaHO 3 / 1 mol CO 2 x 84.0 g NaCO 3 / 1 mol NaHCO 3 = 84.0 g/mol x 0.04926 mol CO 2 = 4.13784 grams Volume of vinegar required (0.833 M acetic acid) Show work, including all units, in the space provided. Volume ( mL ) = 0.04992 molsCO 2 0.833 mol / l x 1 co 2:1 NaCHO 3 x 1000 mL L Volume (mL) = 0.05992 L x 1000mL/L = 59.92 mL vinegar ****If you did not use molar ratios in the calculations for Mass NaHCO 3 and Volume of Vinegar you are completing the calculation incorrectly.****
Data Table 2: Model Air Bag Trial # NaHCO 3 (grams) Vinegar (mL) Observations 1 4.17 g 60mL Bag inflated but still has some room to expand. It was hard to get all the air out before the reaction without mixing the two reactants. 2 4.41 g 65mL Bag does inflate as fast as I would think an airbag would, bag inflated and seems to be alittle bigger than previous experiment. 3 4.99 g 75mL Bag is all the way inflated and inflated quicker, reaction might be quicker the more reactants reacting. Data Table 3: 80-L Driver-Side Air Bag Activity Data and Calculations Moles of CO 2 required to inflate 80L bag at room temperature and pressure. Show work, including all units, in the space provided. 0.9949 atm x 80 L 0.0821 atm L mol K x 295.15 K = 79.592 atm L 24.231815 atm L mol = 3.2846 mol CO 2 Balanced equation for the reaction of NaHCO 3 and CH 3 COOH NaHCO 3 (s)+ CH 3 COOH(aq) H 2 O(l) + CO 2 (g) + C 2 H 3 NaO 2 (aq) Na 1:1 H 5:5 C 3:3 O 5:5 Mass of NaHCO 3 needed for the reaction. Show work, including all units, in the space provided. Grams = 3.2846 mol CO 2 x 1mol NaHO 3 / 1 mol CO 2 x 84.0 g NaCO 3 / 1 mol NaHCO 3 = 84.0 g/mol x 3.2846 mol CO 2 = 275.90 grams NaHCO 3 Volume of vinegar required (0.833 M acetic acid) Show work, including all units, in the space provided. Volume ( mL ) = 3.2846 molsCO 2 0.833 mol / l x 1 co 2:1 NaCHO 3 x 1000 mL L Volume (mL) = 3.943 L x 1000mL/L = 3943 mL vinegar ****If you did not use molar ratios in the calculations for Mass NaHCO 3 and Volume of Vinegar you are completing the calculation incorrectly.**** ©2016 2 Carolina Biological Supply Company
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