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Write a balanced chemical equation for the reaction of solid strontium with iodine gas.

Short Answer

Expert verified
The balanced chemical equation is: \(Sr(s) + I2(g) \rightarrow SrI2(s)\).

Step by step solution

01

Identify the Reactants

First, we need to identify the chemical formulas of the reactants. Strontium is a chemical element with the symbol Sr and iodine gas consists of I2 molecules. So, the reactants are solid strontium (Sr) and iodine gas (I2).
02

Write the Unbalanced Equation

Next, write the unbalanced chemical equation by placing the reactants on the left side and the expected product on the right side. The product of the reaction between a metal and a non-metal is an ionic compound, so the product here will be strontium iodide (SrI2). The unbalanced equation is: \(Sr(s) + I2(g) \rightarrow SrI2(s)\).
03

Balance the Equation

Finally, balance the chemical equation. Strontium, Sr, has a valency of +2, and iodine, I, has a valency of -1. For the ionic compound strontium iodide, we need two iodine atoms for every strontium atom to balance the charges. Also, iodine is diatomic in nature, meaning it exists as I2 in gaseous form. This leads to the balanced equation: \(Sr(s) + I2(g) \rightarrow SrI2(s)\). Since there is one Sr atom on both sides and 2 I atoms on both sides, the equation is balanced as written.

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Key Concepts

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

Chemical Reaction
A chemical reaction is the process where reactants transform into products through the breaking and forming of chemical bonds. Each reaction is governed by certain laws, one of which is the law of conservation of mass. This law states that mass is neither created nor destroyed in a chemical reaction. Therefore, the total mass of reactants must equal the total mass of products.

For example, when solid strontium reacts with iodine gas, they form a new substance, strontium iodide. This transformation is indicative of a chemical reaction occurring. It involves a metal (strontium) reacting with a non-metal (iodine), typically to form an ionic compound. Observing the changes and characterizing the reaction helps chemists understand and predict the behaviors of substances in various conditions.
Stoichiometry
Stoichiometry is essentially the 'mathematics of chemistry.' It deals with the quantitative relationships between the substances as they participate in chemical reactions. The principles of stoichiometry are used to ensure that the Law of Conservation of Mass is respected in the balanced chemical equations.

To apply stoichiometry, as first seen with the reaction of strontium and iodine, we start with the unbalanced equation and adjust coefficients to reflect the mass balance. Initially, Sr and I2 might not have the same number of atoms on both sides, which is a common situation in unbalanced equations. Stoichiometry allows for determining the correct amount of reactants needed to produce desired quantities of products, as well as predicting the outcomes of reactions.
Valency
Valency is an integral part of understanding how elements combine during chemical reactions. It is defined as the combining power of an element, or the number of hydrogen atoms that can combine with or displace to form a compound. In the context of ionic compounds, it reflects the number of electrons an atom can lose, gain, or share to become stable.

Understanding Valency in Ionic Compounds

For ionic compounds, the valency of an element is crucial for determining the formula. The metal strontium (Sr) has a valency of +2, meaning it can donate two electrons. Iodine (I), on the other hand, has a valency of -1 and can accept one electron.

To achieve electrical neutrality, the charges must balance. That is why one Sr atom combines with two I atoms, leading to the formation of SrI2. This valency-driven pairing is what makes stoichiometry possible and allows us to write the balanced chemical equation for the reaction between strontium and iodine.

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Most popular questions from this chapter

Write a balanced chemical equation for each reaction. a. Solid lead(II) sulfide reacts with aqueous hydrobromic acid (HBr) to form solid lead(II) bromide and dihydrogen monosulfide gas. b. Gaseous carbon monoxide reacts with hydrogen gas to form gaseous methane \(\left(\mathrm{CH}_{4}\right)\) and liquid water. c. Aqueous hydrochloric acid (HCl) reacts with solid manganese(IV) oxide to form aqueous manganese(II) chloride, liquid water, and chlorine gas. d. Liquid pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) reacts with gaseous oxygen to form gaseous carbon dioxide and liquid water.

Balance each chemical equation. a. \(\mathrm{Na}_{2} \mathrm{S}(a q)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow \mathrm{NaNO}_{3}(a q)+\mathrm{CuS}(s)\) b. \(\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g)\) c. \(\operatorname{HCl}(a q)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)\) \(\mathrm{d} . \mathrm{FeS}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{S}(g)\)

Classify each property as physical or chemical. $$\begin{array}{l}{\text { a. the boiling point of ethyl alcohol }} \\ {\text { b. the temperature at which dry ice cvaporates }} \\ {\text { c. the tendency of iron to rust }} \\ {\text { d. the color of gold }}\end{array}$$

Consider the balanced equation: $$\mathrm{SiO}_{2}(s)+3 \mathrm{C}(s) \longrightarrow \mathrm{SiC}(s)+2 \mathrm{CO}(g)$$ Complete the table showing the appropriate number of moles of reactants and products. If the number of moles of a reactant is provided, fill in the required amount of the other reactant, as well as the moles of each product that forms. If the number of moles of a product is provided, fill in the required amount of each reactant to make that amount of product, as well as the amount of the other product that is made.

Complete and balance each combustion reaction equation: $$\mathrm {a} .\mathrm{C}_{4} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \longrightarrow \quad \mathrm{b} . \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow$$ $$ \mathrm{c} . \mathrm{CS}_{2}(s)+\mathrm{O}_{2}(g) \longrightarrow \quad \text { d. } \mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}(l)+\mathrm{O}_{2}(g) \longrightarrow $$

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