Question

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

b) Balanced equation for the reaction of NaHCO3 and CH3COOH to CO2

c)Mass of NaHCO3 needed for the reaction (84.0 g/mol)

d) Volume of vinegar required (0.833 M acetic acid)

Answer 1

a. n = 6,54 moles.

b. CH₃COOH(aq) + NaHCO₃(aq) → CH₃COONa(aq) + H₂O (l) + CO₂(g)

c. 549g of NaHCO₃

d. 7,85L of CH₃COOH

Step-by-step explanation:

Answer 2

a. n = 6,54 moles.

b. CH₃COOH(aq) + NaHCO₃(aq) → CH₃COONa(aq) + H₂O (l) + CO₂(g)

c. 549g of NaHCO₃

d. 7,85L of CH₃COOH

Step-by-step explanation:

a. It is possible to know moles of CO₂ required to inflate the air bag using:

n = PV/RT

Where n are moles; P is pressure (1atm at room conditions); V is volume (160L); R is gas constant (0,082 atmL/molK); and T is temperature (298,15K at room conditions.

Replacing:

n = 6,54 moles.

b. The reaction of CH₃COOH with NaHCO₃ produce:

CH₃COOH(aq) + NaHCO₃(aq) → CH₃COONa(aq) + H₂O (l) + CO₂(g)

c. 1 mol of CO₂ is produced from 1 mol of NaHCO₃, that means 6,54moles of CO₂ are produced from 6,54 moles of NaHCO₃. In grams:

6,54 moles NaHCO₃×
`(84,0g)/(1mol)` = 549g of NaHCO₃

d. Again, you require 6,54 moles of CH₃COOH. If your acetic acid solution is 0,833M you need:

6,54moles×
`(1L)/(0,833mol)` = 7,85L of CH₃COOH

I hope it helps!