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tangare [24]
3 years ago
14

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

Chemistry
1 answer:
sladkih [1.3K]3 years ago
3 0

Answer:

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

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:

<em>n = 6,54 moles.</em>

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 <em>6,54 moles of NaHCO₃</em>. In grams:

6,54 moles NaHCO₃×\frac{84,0g}{1mol} = <em>549g of NaHCO₃</em>

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

6,54moles×\frac{1L}{0,833mol} = <em>7,85L of CH₃COOH</em>

<em></em>

I hope it helps!

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The expression of K_p is given by:

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