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Chapter 9: Problem 76

Consider the balanced equation $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ What mole ratio enables you to calculate the number of moles of oxygen needed to react exactly with a given number of moles of \(\mathrm{C}_{3} \mathrm{H}_{8}(g) ?\) What mole ratios enable you to calculate how many moles of each product form from a given number of moles of \(\mathrm{C}_{3} \mathrm{H}_{8} ?\)

Short Answer

Expert verified
To calculate the number of moles of oxygen required to react exactly with a given number of moles of C₃H₈, use the mole ratio of O₂ to C₃H₈, which is 5:1. For determining how many moles of each product form from a given number of moles of C₃H₈, use the mole ratios of CO₂ to C₃H₈ (3:1) and H₂O to C₃H₈ (4:1).

Step by step solution

01

Identify the balanced chemical equation

First, we have the balanced chemical equation as follows: \[ \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 3\mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \]
02

Determine the mole ratio of oxygen to propane

From the balanced equation, we can see that 1 mole of C₃H₈ reacts with 5 moles of O₂. Therefore, the mole ratio of O₂ to C₃H₈ is 5:1.
03

Determine the mole ratio of CO₂ and H₂O to propane

From the balanced equation, we can see that 1 mole of C₃H₈ forms 3 moles of CO₂ and 4 moles of H₂O. Therefore, the mole ratios of CO₂ to C₃H₈ is 3:1, and the mole ratio of H₂O to C₃H₈ is 4:1.
04

Answer the exercise questions

To find the number of moles of oxygen needed to react exactly with a given number of moles of C₃H₈, use the mole ratio of O₂ to C₃H₈, which is 5:1. To find how many moles of each product form from a given number of moles of C₃H₈, use the mole ratios of CO₂ to C₃H₈, which is 3:1, and the mole ratio of H₂O to C₃H₈, which is 4:1.

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Key Concepts

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

Mole Ratio
In stoichiometry, the mole ratio is a crucial concept that helps us make sense of the relationships in a chemical reaction. The mole ratio shows the relationship between the amounts of reactants used and products formed, expressed in moles.
  • For any chemical reaction, mole ratios are derived from the coefficients of substances in a balanced chemical equation.
  • A simple example is found in the combustion of propane, where the balanced equation tells us that 1 mole of propane (\(\mathrm{C}_{3} \mathrm{H}_{8}\)) reacts with 5 moles of oxygen (\(\mathrm{O}_{2}\)).
To calculate how much oxygen is needed for a certain amount of propane, or how much of a product is formed, these mole ratios are used. If you start with 2 moles of propane, based on the mole ratio of 5:1, you'll need 10 moles of oxygen to completely combust the propane.
Balanced Chemical Equation
A balanced chemical equation is essential for solving stoichiometry problems because it ensures the law of conservation of mass is met. This law states that matter cannot be created or destroyed, only transformed.In a balanced equation, the number of atoms for each element is the same on both sides (reactants and products) of the reaction. The balanced propane combustion equation is:\[\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\]Here, propane and oxygen on one side, react completely to form carbon dioxide and water on the other. The coefficients (1 for propane, 5 for oxygen, 3 for carbon dioxide, and 4 for water) tell us how many moles are involved in the reaction, maintaining balance.
Chemical Reactions
Chemical reactions describe the transformation of substances through the breaking and forming of bonds, resulting in new substances. This is shown through chemical equations.
  • Reactants (starting substances) transform into products (substances formed).
  • Chemical equations depict the reactants and products as well as their proportions, thanks to mole ratios and balancing.
For example, in the reaction of propane combustion, the reactants propane and oxygen undergo a reaction to form carbon dioxide and water, substances different in structure and properties from the reactants. Understanding chemical reactions involves knowing both the starting materials and the results, showing how mass and energy are transferred.
Propane Combustion
Combustion of propane is a specific type of chemical reaction known as an exothermic reaction—one that releases energy in the form of heat and light.Propane (\(\mathrm{C}_{3} \mathrm{H}_{8}\)), when exposed to oxygen, combusts to form carbon dioxide (\(\mathrm{CO}_{2}\)) and water (\(\mathrm{H}_{2} \mathrm{O}\)), releasing a significant amount of energy. This process is important for heating and cooking applications.
  • The balanced equation for propane combustion is a model for understanding the conversion of propane and oxygen into products.
  • This combustion demonstrates principles of balanced equations and mole ratios, ensuring accurate calculations for practical uses like fuel efficiency.
By capturing energy from the reaction, systems can efficiently use the chemical potential energy stored in hydrocarbons like propane.

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Most popular questions from this chapter

For each of the following incomplete and unbalanced equations, indicate how many moles of the second reactant would be required to react completely with 0.145 mol of the first reactant. a. \(\mathrm{BaCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow\) b. \(\operatorname{AgNO}_{3}(a q)+\operatorname{NaCl}(a q) \rightarrow\) c. \(\operatorname{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \rightarrow\) d. \(C_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \rightarrow\)

Silicon carbide, \(\mathrm{SiC},\) is one of the hardest materials known. Surpassed in hardness only by diamond, it is sometimes known commercially as carborundum. Silicon carbide is used primarily as an abrasive for sandpaper and is manufactured by heating common sand (silicon dioxide, \(\mathrm{SiO}_{2}\) ) with carbon in a furnace. $$\mathrm{SiO}_{2}(s)+\mathrm{C}(s) \rightarrow \mathrm{CO}(g)+\mathrm{SiC}(s)$$ What mass of silicon carbide should result when 1.0 kg of pure sand is heated with an excess of carbon?

The traditional method of analysis for the amount of chloride ion present in a sample was to dissolve the sample in water and then slowly to add a solution of silver nitrate. Silver chloride is very insoluble in water, and by adding a slight excess of silver nitrate, it is possible effectively to remove all chloride ion from the sample. $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{AgCl}(s) $$ Suppose a \(1.054-\mathrm{g}\) sample is known to contain \(10.3 \%\) chloride ion by mass. What mass of silver nitrate must be used to completely precipitate the chloride ion from the sample? What mass of silver chloride will be obtained?

What is the limiting reactant for a process? Why does a reaction stop when the limiting reactant is consumed, even though there may be plenty of the other reactants present?

For each of the following reactions, give the balanced equation for the reaction and state the meaning of the equation in terms of numbers of individual molecules and in terms of moles of molecules. a. \(\mathrm{UO}_{2}(s)+\mathrm{HF}(a q) \rightarrow \mathrm{UF}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) b. \(\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) c. \(\operatorname{Mg}(s)+\operatorname{HCl}(a q) \rightarrow \operatorname{MgCl}_{2}(a q)+\mathrm{H}_{2}(g)\) d. \(\mathrm{B}_{2} \mathrm{O}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{B}(\mathrm{OH})_{3}(a q)\)
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