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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 COâ‚‚ and Hâ‚‚O 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

Thionyl chloride, \(\mathrm{SOCl}_{2}\), is used as a very powerful drying agent in many synthetic chemistry experiments in which the presence of even small amounts of water would be detrimental. The unbalanced chemical equation is $$\mathrm{SOCl}_{2}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{HCl}(g)$$ Calculate the mass of water consumed by complete reaction of \(35.0 \mathrm{g}\) of \(\mathrm{SOCl}_{2}\).

For each of the following unbalanced chemical equations, suppose that exactly \(5.00 \mathrm{g}\) of each reactant is taken. Determine which reactant is limiting, and calculate what mass of each product is expected (assuming that the limiting reactant is completely consumed). a. \(\mathrm{S}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) b. \(\operatorname{MnO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \rightarrow \mathrm{Mn}\left(\mathrm{SO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) c. \(\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) d. \(\mathrm{AgNO}_{3}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{Ag}(s)+\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)\)

For each of the following balanced reactions, calculate how many moles of product would be produced by complete conversion of 0.15 mol of the reactant indicated in boldface. State clearly the mole ratio used for the conversion. a. \(2 \mathbf{M} g(s)+O_{2}(g) \rightarrow 2 M g O(s)\) b. \(2 \mathrm{Mg}(s)+\mathbf{O}_{2}(g) \rightarrow 2 \mathrm{MgO}(s)\) c. 4 Fe \((s)+3 O_{2}(g) \rightarrow 2 F e_{2} O_{3}(s)\) d. \(4 \mathrm{Fe}(s)+3 \mathbf{O}_{2}(g) \rightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\)

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?

If sodium peroxide is added to water, elemental oxygen gas is generated: $$\mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{NaOH}(a q)+\mathrm{O}_{2}(g)$$ Suppose \(3.25 \mathrm{g}\) of sodium peroxide is added to a large excess of water. What mass of oxygen gas will be produced?

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