Question

An air bag is deployed by utilizing the following reaction (the nitrogen gas produced inflates the air bag): $$ 2 \mathrm{NaN}_{3}(s) \rightarrow 2 \mathrm{Na}(s)+3 \mathrm{~N}_{2}(g) $$ If \(10.5 \mathrm{~g}\) of \(\mathrm{NaN}_{3}\) is decomposed, what theoretical mass of sodium should be produced? If only \(2.84 \mathrm{~g}\) of sodium is actually collected, what is the percent yicld?

Short answer

The theoretical mass of sodium produced from 10.5 g of \(\mathrm{NaN}_{3}\) is approximately 3.715 g, and the percent yield of the reaction is approximately 76.43%.

Step-by-step solution

Write the balanced chemical equation

We are given the balanced chemical equation for the decomposition reaction: \(2 \mathrm{NaN}_{3}(s) \rightarrow 2 \mathrm{Na}(s)+3 \mathrm{N}_{2}(g)\)

Calculate molar masses of \(\mathrm{NaN}_{3}\) and \(\mathrm{Na}\)

Let's find the molar masses of \(\mathrm{NaN}_{3}\) and \(\mathrm{Na}\). For \(\mathrm{NaN}_{3}\): Molar mass = (1 * 22.99) + (3 * 14.01) = 22.99 + 42.03 = 65.02 g/mol For \(\mathrm{Na}\): Molar mass = 22.99 g/mol

Calculate moles of \(\mathrm{NaN}_{3}\)

Given 10.5 g of \(\mathrm{NaN}_{3}\), we need to convert that into moles. To do this, divide the given mass by the molar mass of \(\mathrm{NaN}_{3}\): Moles of \(\mathrm{NaN}_{3}\) = (10.5 g) / (65.02 g/mol) = 0.1616 mol (approximately)

Use stoichiometry to calculate moles of \(\mathrm{Na}\)

From the balanced chemical equation, 2 moles of \(\mathrm{NaN}_{3}\) produce 2 moles of \(\mathrm{Na}\). Using stoichiometry, we can now calculate the amount of \(\mathrm{Na}\) produced: Moles of \(\mathrm{Na}\) = (0.1616 mol of \(\mathrm{NaN}_{3}\)) * (2 mol of \(\mathrm{Na}\) / 2 mol of \(\mathrm{NaN}_{3}\)) = 0.1616 mol (approximately)

Convert moles of \(\mathrm{Na}\) to grams (theoretical mass)

Now, we can find the theoretical mass of \(\mathrm{Na}\) produced by multiplying the amount of moles by its molar mass: Theoretical mass of \(\mathrm{Na}\) = (0.1616 mol) * (22.99 g/mol) = 3.715 g (approximately)

Calculate percent yield

The actual amount of sodium collected is given as 2.84 g. To find the percent yield, we use the formula: Percent yield = (actual mass / theoretical mass) * 100 % Percent yield = (2.84 g / 3.715 g) * 100 % = 76.43 % (approximately) Answer: The theoretical mass of sodium produced from 10.5 g of \(\mathrm{NaN}_{3}\) is approximately 3.715 g, and the percent yield of the reaction is approximately 76.43%.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the quantitative relationships between reactants and products in a chemical reaction. It's essentially a bridge between the world of atoms and the world we can measure, like grams and liters. In a stoichiometric calculation, you typically start with a balanced chemical equation, which tells you the proportions in which substances react or are produced.

For example, in the decomposition of sodium azide (\( ext{NaN}_3 \)), the stoichiometric coefficients tell us that 2 moles of \( ext{NaN}_3 \) decompose to produce 2 moles of \( ext{Na} \). This means that for every mole of \( ext{NaN}_3 \), there is a mole of \( ext{Na} \) formed. Thus, stoichiometry helps in predicting how much product can be formed from a given amount of reactant.

Important points about stoichiometry:

• Balancing the chemical equation is crucial.
• The coefficients from the balanced equation are used to establish mole ratios.
• Mole ratios are used to convert between substances in a reaction.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. They are characterized by changes in chemical properties and energy. A chemical equation represents this change and is written with reactants on the left side and products on the right.

Consider the decomposition reaction of sodium azide, where \(2 ext{NaN}_3\) decomposes into \(2 ext{Na}\) and \(3 ext{N}_2\). This is a classic example of a chemical change where new substances with different properties are formed. Each chemical equation must respect the law of conservation of mass, which means the same number of each type of atom must be present on both sides of the equation.

Features of chemical reactions:

• They involve breaking old bonds and forming new ones.
• They can result in energy release or absorption (exothermic or endothermic).
• The stoichiometry of a reaction gives the proportionate amount of reactants and products.

Percent Yield

Percent yield is a valuable concept in chemical reactions, indicating the efficiency of a reaction. It compares the actual yield (what was obtained from the experiment) to the theoretical yield (what was expected according to stoichiometric calculations).

The formula for percent yield is:
\[ \text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100\% \]

In the example given, the actual mass of sodium obtained from the reaction was 2.84 g, while the theoretical mass was calculated to be 3.715 g. Using the percent yield formula, we find the percent yield to be approximately 76.43%.

Why is percent yield important?

• It helps in assessing how efficient the reaction was.
• Economic decisions in industrial processes often depend on yield.
• Understanding yield shortfall can guide improvements in experimental methods.