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The vigorous reaction between aluminum and iodine gives the balanced equation: $$ 2 \mathrm{Al}(s)+3 \mathrm{I}_{2}(s) \rightarrow 2 \mathrm{AlI}_{3}(s) $$ What do the coefficicnts in this balanced chemical cquation tell us about the proportions in which these substances react on a macroscopic (mole) basis?

Short Answer

Expert verified
In the balanced chemical equation \( 2 \mathrm{Al}(s)+3 \mathrm{I}_{2}(s) \rightarrow 2 \mathrm{AlI}_{3}(s) \), the coefficients represent the mole ratio of the reactants and product. They tell us that 2 moles of aluminum (Al) react with 3 moles of iodine (I₂) to produce 2 moles of aluminum iodide (AlI₃). These coefficients give us the necessary proportions for the reactants and product on a macroscopic (mole) basis.

Step by step solution

01

Identify the coefficients in the balanced equation

The balanced chemical equation is: \( 2 \mathrm{Al}(s)+3 \mathrm{I}_{2}(s) \rightarrow 2 \mathrm{AlI}_{3}(s) \) The coefficients in this equation are: - 2 for aluminum (Al) - 3 for iodine (I₂) - 2 for aluminum iodide (AlI₃)
02

Interpret the coefficients as a mole ratio

The coefficients in the balanced chemical equation represent the mole ratio of the reactants and the product. In this case, the coefficients tell us that: - 2 moles of aluminum (Al) react with - 3 moles of iodine (I₂) to produce - 2 moles of aluminum iodide (AlI₃)
03

Understand the proportions

The coefficients indicate how many moles of each substance are required for the reaction to occur in a proportional manner. In this case, the proportions are as follows: - For every 2 moles of aluminum (Al) reacting, there must be 3 moles of iodine (I₂) present. - Similarly, for every 3 moles of iodine (I₂) reacting, there must be 2 moles of aluminum (Al) present. - The reaction will produce 2 moles of aluminum iodide (AlI₃) for every 2 moles of aluminum and 3 moles of iodine consumed. In conclusion, the coefficients in the balanced chemical equation tell us about the proportions of the reactants and product on a macroscopic (mole) basis. In this particular reaction, 2 moles of aluminum react with 3 moles of iodine to form 2 moles of aluminum iodide.

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Key Concepts

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

Mole Ratio
The mole ratio in a chemical reaction is a crucial concept that helps us understand the proportions in which substances react or form. It is derived from the coefficients of a balanced chemical equation and gives us insight into the quantitative relationship between reactants and products.

In the given reaction between aluminum and iodine, the balanced equation is:\[2 \mathrm{Al}(s)+3 \mathrm{I}_{2}(s) \rightarrow 2 \mathrm{AlI}_{3}(s)\]
The mole ratio here is
  • 2 moles of aluminum (Al)
  • 3 moles of iodine (Iâ‚‚)
  • 2 moles of aluminum iodide (AlI₃)
This means that to completely react 2 moles of aluminum, you need exactly 3 moles of iodine. In turn, this combination produces 2 moles of aluminum iodide. Understanding the mole ratio is essential for scaling reactions up or down, and for calculating how much of a reactant is needed or how much product is expected.
Balanced Chemical Equation
A balanced chemical equation ensures that the law of conservation of mass is upheld. This law states that matter cannot be created or destroyed in a chemical reaction. Therefore, a balanced chemical equation has the same number of atoms of each element on both sides of the equation.

The balanced equation for the reaction between aluminum and iodine is \[2 \mathrm{Al}(s)+3 \mathrm{I}_{2}(s) \rightarrow 2 \mathrm{AlI}_{3}(s)\]
Here, the equation is balanced because:
  • There are 2 Al atoms on the reactant side and 2 Al atoms in the product, maintaining the balance of aluminum atoms.
  • Iodine, in the reactant form Iâ‚‚ (which means each molecule contains 2 iodine atoms), has 3 molecules contributing to a total of 6 iodine atoms, which matches the 6 iodine atoms present in 2 molecules of the product AlI₃.
Balancing chemical equations is a fundamental skill in chemistry, needed to accurately describe a chemical reaction.
Chemical Reactions
Chemical reactions are processes where substances, known as reactants, are transformed into new substances, called products. Understanding chemical reactions is pivotal, as it forms the basis of studying how materials interact, change, and bond.

In the reaction at hand, aluminum (\(\mathrm{Al}\)) reacts with iodine (\(\mathrm{I}_2\)) to form aluminum iodide (\(\mathrm{AlI}_3\)). This reaction can be characterized by several key features:
  • Reactants: Substances that start the reaction, such as aluminum and iodine in this case.
  • Products: New substances formed as a result of the reaction, here, aluminum iodide.
  • Reaction Type: This particular reaction is a synthesis reaction, where two or more reactants combine to form a single product.
Chemical reactions encompass energy changes and may involve breaking or forming of chemical bonds, leading to the transformation of substances. Understanding the stoichiometry of the reaction helps predict the quantities needed and allows for manipulating conditions to achieve the desired outcome efficiently.

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Most popular questions from this chapter

For the balanced chemical equation for the combination reaction of hydrogen gas and oxygen gas $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ explain why we know that \(2 \mathrm{~g}\) of \(\mathrm{H}_{2}\) reacting with \(1 \mathrm{~g}\) of \(\mathrm{O}_{2}\) will not result in the production of \(2 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\)

For cach of the following balanced reactions, calculate how many moles of each product would be produced by complete conversion of 0.50 mole of the reactant indicated in boldface. Indicate clearly the mole ratio used for the conversion. a. \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\) b. \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) c. \(2 \mathrm{Al}(s)+6 \mathrm{HCl}(a q) \rightarrow 2 \mathrm{AlCl}_{3}(a q)+3 \mathrm{H}_{2}(g)\) d. \(\mathbf{C}_{3} \mathbf{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\)

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?

Consider the following unbalanced chemical equation. $$ \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ If \(67.4 \mathrm{~g}\) of lithium hydroxide reacts with excess carbon dioxide, what mass of lithium carbonate will be produced?

Although we usually think of substances as "burning" only in oxygen gas, the process of rapid oxidation to produce a flame may also take place in other strongly oxidiring gases. For cxample, when iron is heated and placed in pure chlorine gas, the iron "burns" according to the following (unbalanced) reaction: $$ \mathrm{Fe}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{FeCl}_{3}(s) $$ How many milligrams of iron(III) chloride result when \(15.5 \mathrm{mg}\) of iron is reacted with an excess of chlorine gas?

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