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Chapter 9: Problem 9

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?

Short Answer

Expert verified
To calculate the moles of oxygen gas needed to react completely with a given number of moles of aluminum metal, use the mole ratio \(\frac{3 \:moles \:of\: O_2}{4 \:moles \:of\: Al}\). To calculate the number of moles of aluminum oxide expected if a given number of moles of aluminum metal reacts completely, use the mole ratio \(\frac{1 \:mole \:of\: Al_2O_3}{2\: moles \:of\: Al}\).

Step by step solution

01

Identify the coefficients in the balanced chemical equation.

From the balanced chemical equation, we can see that the coefficients are: - 4 for aluminum (Al) - 3 for oxygen gas (O2) - 2 for aluminum oxide (Al2O3)
02

Find the mole ratio for moles of oxygen gas needed for the given moles of aluminum metal.

To find out how many moles of oxygen gas (O2) would be needed to react completely with given moles of aluminum (Al), we look at the coefficient ratio between Al and O2. The ratio is 4:3. So for every 4 moles of Al, 3 moles of O2 are required. Therefore, the mole ratio for Al to O2 is: $$ \frac{3 \:moles \:of\: O_2}{4 \:moles \:of\: Al} $$
03

Find the mole ratio for the number of moles of aluminum oxide expected for the given moles of aluminum metal.

To find out how many moles of aluminum oxide (Al2O3) would be expected if a given number of moles of aluminum (Al) reacts completely, we look at the coefficient ratio between Al and Al2O3. The ratio is 4:2 or simplified to 2:1. So for every 2 moles of Al, 1 mole of Al2O3 is formed. Therefore, the mole ratio for Al to Al2O3 is: $$ \frac{1 \:mole \:of\: Al_2O_3}{2\: moles \:of\: Al} $$

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Key Concepts

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

Stoichiometry Made Simple
Stoichiometry is a fundamental concept in chemistry that helps us quantify the relationships between reactants and products in chemical reactions.
It might sound complicated, but it's essentially about using ratios derived from balanced equations to make predictions about the amounts of substances involved in a reaction.
This approach allows scientists to calculate how much of a reactant is needed, or how much product will be formed, once they know the quantities of other substances involved.
In the exercise provided, stoichiometry helps us determine the amounts of oxygen gas needed to react with aluminum and the amount of aluminum oxide produced.
By using mole ratios, which are derived from the coefficients in the balanced chemical equation, we can ensure that our calculations are accurate and aligned with the conservation of mass in chemical reactions.
To practice stoichiometry, one must become comfortable finding the mole ratios and applying them to calculate quantities in chemical reactions.
Understanding Balanced Chemical Equations
Balanced chemical equations are the foundation of stoichiometry.
A balanced equation has the same number of each type of atom on both sides of the equation, which reflects the conservation of mass: matter cannot be created or destroyed.
When balancing equations, we adjust the coefficients, which are the numbers in front of molecules or atoms, not the subscripts in the chemical formulas.
In the exercise equation \(4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3\), the numbers 4, 3, and 2 are coefficients, showing the stoichiometric relationships of reactants and products.
These coefficients tell us that:
  • 4 moles of aluminum react with 3 moles of oxygen gas
  • to form 2 moles of aluminum oxide.
The balanced equation ensures we have the correct mole ratios needed for stoichiometric calculations, making it an essential tool for predicting the outcomes of chemical reactions.
Chemical Reactions and Their Participants
Chemical reactions involve the transformation of reactants into products, and they can take many forms, from combustion to synthesis or decomposition.
In the context of the exercise, we are looking at a synthesis reaction where aluminum reacts with oxygen to form aluminum oxide.
Understanding the basic nature of a chemical reaction helps in identifying the species involved by recognizing their chemical symbols and phases.
The chemical equation \(4 \text{Al}(s) + 3 \text{O}_2(g) \rightarrow 2 \text{Al}_2\text{O}_3(s)\) shows:
  • Aluminum (Al) starting as a solid metal
  • Oxygen (\(\text{O}_2\)) as a gaseous reactant
  • And forming aluminum oxide (\(\text{Al}_2\text{O}_3\)), a solid compound.
Recognizing the states of matter and the type of reaction can guide you in making sense of the process and understanding the transformations that occur during the reaction.

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Most popular questions from this chapter

Consider the following reaction: \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) If a container were to have only 10 molecules of \(\mathrm{O}_{2}(g)\) and 10 molecules of \(\mathrm{NH}_{3}(g),\) how many total molecules (reactant and product) would be present in the container after the above reaction goes to completion? b. Using "microscopic" pictures, draw the total molecules present inside the container after the reaction occurs. c. What mass of \(\mathrm{NO}(g)\) is present in the container after the reaction occurs? (Report your final answer to 4 significant figures.

Although they were formerly called the inert gases, at least the heavier elements of Group 8 do form relatively stable compounds. For example, xenon combines directly with elemental fluorine at elevated temperatures in the presence of a nickel catalyst. $$ \mathrm{Xe}(g)+2 \mathrm{~F}_{2}(g) \rightarrow \mathrm{XeF}_{4}(s) $$ What is the theoretical mass of xenon tetrafluoride that should form when \(130 .\) g of xenon is reacted with 100.8 of \(\mathrm{F}_{2} ?\) What is the percent yield if only \(145 \mathrm{~g}\) of \(\mathrm{XeF}_{4}\) is actually isolated?

When elemental carbon is burned in the open atmosphere, with plenty of oxygen gas prescnt, the product is carbon dioxide. $$ \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) $$ However, when the amount of oxygen present during the burning of the carbon is restricted, carbon monoxide is more likely to result. $$ 2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}(g) $$ What mass of each product is expected when a \(5.00-\mathrm{g}\) sample of pure carbon is burned under cach of these conditions?

One process for the commercial production of baking soda (sodium hydrogen carbonate) involves the following reaction, in which the carbon dioxide is used in its solid form ("dry ice" enough for the sodium hydrogen carbonate to precipitate: $$ \mathrm{NaCl}(a q)+\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(s) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{NaHCO}_{3}(s) $$ Because they are relatively cheap, sodium chloride and water are typically present in excess. What is the expected yield of \(\mathrm{NaHCO}_{3}\) when one performs such a synthesis using \(10.0 \mathrm{~g}\) of ammonia and \(15.0 \mathrm{~g}\) of dry ice, with an excess of \(\mathrm{NaCl}\) and water?

For each of the following tunbalanced equations, calculate how many moles of the second reactant would be required to react completely with 0.557 grams of the first reactant. a. \(\mathrm{Al}(s)+\mathrm{Br}_{2}(l) \rightarrow \mathrm{AlBr}_{3}(s)\) b. \(\mathrm{Hg}(s)+\mathrm{HClO}_{4}(a q) \rightarrow \mathrm{Hg}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{H}_{2}(g)\) c. \(\mathrm{K}(s)+\mathrm{P}(s) \rightarrow \mathrm{K}_{3} \mathrm{P}(s)\) d. \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(l)+\mathrm{HCl}(g)\)
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