Introduction

Airbags and Efficiency

Airbags are important safety devices used in automobiles to reduce the chance of injury during a collision.

Throughout this experiment you will be experimenting with the chemistry of the following reaction:

N A H C O three, solid, plus C H 3 C O O H, aqueous, yield C H 3 C O O N A, aqueous, plus C O 2, gas, plus H 2 O, liquid

The ultimate goal of your experiments is to create the most efficient airbag possible.

With regards to efficiency, an ideal airbag would use the least amount of materials and be as light as possible while still being able to inflate to a proper volume so that it prevents injuries.

There are many ways to measure efficiency, but for this particular experiment we will focus on two, volume to mass ratio and atom economy.

The volume to mass ratio is simply the volume of your inflated airbag divided by the total mass of the airbag.

Volume to mass ratio equals volume of inflated airbag divided by mass of airbag

The atom economy is the mass of the desired product divided by the total mass of all products and byproducts.

Atom economy equals mass of desired products divided by the quantity total mass of products plus by products

Stoichiometry and Limiting Reagents

During this experiment, you will be asked to create different chemical ratios of sodium bicarbonate (NaHCO₃) and acetic acid (CH₃COOH), and calculate the theoretical volume of carbon dioxide, CO₂, produced. This will require both stoichiometry and limiting reagents to do the calculation properly.

Stoichiometry is used to convert between the moles of one species and the moles of another using a balanced chemical equation.

Six N A N three, solid, plus F E two O three, solid, yield nine N two, gas, plus three N A two O, solid, plus 2 F E, solid.

Using the equation above, we can see that for every 1 mole of iron (III) oxide, Fe₂O₃, we would produce 9 moles of nitrogen, N₂.

Example 1: How many moles of nitrogen, N₂, will be produced by reacting 6.0 g of iron (III) oxide, Fe₂O₃, with a stoichiometrically equivalent amount of sodium azide, NaN₃?

6.0 grams F E two O three, times 1 mole F E two O three over 159.69 grams F E two O three, times 9 moles N two over 1 mole F E two O three, equals 0.34 moles of N two.

Answer: 0.34 moles of N₂

This formula tells us exactly how many moles of products will be produced provided the ratios of the reactants are exactly what is shown in the formula (stoichiometric amounts).

However, sometimes the quantities of our reactants are not in exact stoichiometric amounts and we have a shortage of one reactant and an excess of the other. Sometimes there are advantages to performing reactions this way.

When this happens, the reactant that is in short supply is called the limiting reagent and the reaction can only continue making product until this reactant is used up, leaving behind excess of the other reactants.

Example 2: How many moles of nitrogen, N₂, will be produced by reacting 6.0 g of iron (III) oxide, Fe₂O₃, with 13 g of sodium azide, NaN₃?

6.0 grams F E 2 O 3, times 1 mole F E 2O 3 over 159.69 grams F E 2 O 3, times 9 moles N 2 over 1 mole F E 2 O 3, equals 0.34 moles of N 2

13 grams N A N 3, times 1 mole N A N 3 over 65.01 grams N A N 3, times 9 moles N 2 over 6 moles N A N 3, equals 0.30 moles of N 2

Answer: Since the sodium azide would produce a smaller amount of N₂, sodium azide is the limiting reagent because it would run out first. This means that we can produce a maximum of 0.30 moles of N₂ with these quantities of reactants.

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