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Chapter 9: Problem 62

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

Expert verified
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

01

Write the balanced chemical equation

We are given the balanced chemical equation for the decomposition reaction: \(2 \mathrm{NaN}_{3}(s) \rightarrow 2 \mathrm{Na}(s)+3 \mathrm{N}_{2}(g)\)
02

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
03

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)
04

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)
05

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)
06

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%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Natural waters often contain relatively high levels of calcium ion, \(\mathrm{Ca}^{2+},\) and hydrogen carbonate ion (bicarbonate), \(\mathrm{HCO}_{3}^{-}\), from the leaching of minerals into the water. When such water is used commercially or in the home, heating of the water leads to the formation of solid calcium carbonate, \(\mathrm{CaCO}_{3}\), which forms a deposit ("scale") on the interior of boilers, pipes, and other plumbing fixtures. $$ \mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(a q) \rightarrow \mathrm{CaCO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If a sample of well water contains \(2.0 \times 10^{-3} \mathrm{mg}\) of \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) per milliliter, what mass of \(\mathrm{CaCO}_{3}\) scale would \(1.0 \mathrm{~mL}\) of this water be capable of depositing?

Consider the balanced chemical equation $$ 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ What mole ratio would you use to calculate how many moles of oxygen gas would be needed to react completely with a given number of moles of aluminum metal? What mole ratio would you use to calculate the number of moles of product that would be expected if a given number of moles of aluminum metal reacts completely?

For each of the following balanced equations, indicate how many moles of the product could be produced by complete reaction of \(1.00 \mathrm{~g}\) of the reactant indicated in boldface. Indicate clearly the mole ratio used for the conversion. a. \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) b. \(\mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CaCO}_{3}(s)\) $$ \text { c. } 4 \mathrm{Na}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Na}_{2} \mathrm{O}(s) $$ d. \(2 \mathbf{P}(s)+3 C l_{2}(g) \rightarrow 2 \mathrm{PCl}_{3}(l)\)

"Smelling salts," which are used to revive someone who has fainted, typically contain ammonium carbonate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\). Ammonium carbonate decomposes readily to form ammonia, carbon dioxide, and water. The strong odor of the ammonia usually restores consciousness in the person who has fainted. The unbalanced equation is $$ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Calculate the mass of ammonia gas that is produced if \(1.25 \mathrm{~g}\) of ammonium carbonate decomposes completcly.

Ammonium nitrate has been used as a high explosive because it is unstable and decomposes into several gaseous substances. The rapid expansion of the gaseous substances produces the explosive force. $$ \mathrm{NH}_{4} \mathrm{NO}_{3}(s) \rightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Calculate the mass of each product gas if \(1.25 \mathrm{~g}\) of ammonium nitrate reacts.
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