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Ch.5 - Introduction to Solutions and Aqueous Solutions
Tro - Chemistry: A Molecular Approach 5th Edition
Tro5th EditionChemistry: A Molecular ApproachISBN: 9780134874371Not the one you use?Change textbook
All textbooksTro 5th EditionCh.5 - Introduction to Solutions and Aqueous SolutionsProblem 44a
Chapter 5, Problem 44a

Complete and balance each equation. If no reaction occurs, write 'NO REACTION.' a. NaNO3(aq) + KCl(aq) →

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Identify the reactants: NaNO_3 (aq) and KCl (aq).
Determine the possible products by exchanging the anions and cations: NaCl and KNO_3.
Write the unbalanced equation: NaNO_3 (aq) + KCl (aq) → NaCl (aq) + KNO_3 (aq).
Check the solubility of the products: Both NaCl and KNO_3 are soluble in water.
Since all reactants and products are soluble, no precipitate forms, and the reaction does not proceed. Therefore, the result is 'NO REACTION.'

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Double Displacement Reaction

A double displacement reaction, also known as a double replacement reaction, occurs when two compounds exchange ions or bonds to form two new compounds. In this type of reaction, the cations and anions of the reactants switch partners, which can lead to the formation of a precipitate, gas, or a weak electrolyte. Understanding this concept is crucial for predicting the products of the reaction between NaNO3 and KCl.
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Solubility Rules

Solubility rules are guidelines that help predict whether a compound will dissolve in water. For example, most nitrates (like NaNO3) and chlorides (like KCl) are soluble in water, while some combinations may lead to the formation of insoluble products. Knowing these rules is essential for determining if a reaction will occur and if any precipitate will form in the given equation.
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Balancing Chemical Equations

Balancing chemical equations involves ensuring that the number of atoms for each element is the same on both sides of the equation, adhering to the law of conservation of mass. This process often requires adjusting coefficients in front of compounds rather than changing subscripts. Mastery of this concept is necessary to accurately represent the reaction and its products in the equation provided.
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