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Write a balanced equation for reaction of the basic oxide, magnesium oxide, with water.

Short Answer

Expert verified
The balanced equation is \( \text{MgO} + \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2 \).

Step by step solution

01

Identify the Reactants

The problem requires us to identify the reactants involved in the chemical reaction. For this particular reaction, the reactants are magnesium oxide, which is represented as \( \text{MgO} \), and water, which is represented as \( \text{H}_2\text{O} \).
02

Predict the Products

The reaction of magnesium oxide (a basic oxide) with water typically results in the formation of a metal hydroxide. In this case, the metal hydroxide formed is magnesium hydroxide \( \text{Mg(OH)}_2 \).
03

Write the Skeleton Equation

Using the reactants and the predicted product, we can write the skeleton equation for the reaction: \[ \text{MgO} + \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2 \]
04

Balance the Chemical Equation

Next, we need to balance the equation. Counting the atoms on each side of the equation shows that there is 1 atom of Mg, 2 atoms of O, and 2 atoms of H on both sides already. Thus, the equation is balanced as \[ \text{MgO} + \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2 \].

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Key Concepts

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

Magnesium Oxide Reaction
When magnesium oxide (\(\text{MgO}\)) reacts with water (\(\text{H}_2\text{O}\)), it is a delightful example of chemistry in action! This is a type of reaction where a basic oxide joins forces with water. Magnesium oxide is a solid white powder at room temperature and is known as a basic oxide because it reacts with water to form a base rather than an acid.
Metal oxides like magnesium oxide can pair with water molecules due to their chemical composition. The reaction between these two substances produces magnesium hydroxide (\(\text{Mg(OH)}_2\)), a compound that is a weak base. This particular reaction can naturally occur when magnesium burns and forms magnesium oxide, which then reacts with water vapor in the air.
The reaction is not only fascinating but also has practical applications. Magnesium hydroxide is often used as an antacid and in the treatment of wastewater, showcasing how simple chemical reactions can have important real-world uses!
Chemical Equation Balancing
Balancing a chemical equation is like solving a puzzle where the goal is to have the same number of each type of atom on each side of the equation. The law of conservation of mass tells us that mass can't be created or destroyed in a chemical reaction, so we must account for all atoms involved.
In our magnesium oxide reaction, the unbalanced equation looked like this: \(\text{MgO} + \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2\). At first glance, it may seem balanced, but we need to confirm by counting the atoms:
  • 1 magnesium (Mg) atom on both sides
  • 2 oxygen (O) atoms on both sides
  • 2 hydrogen (H) atoms on both sides
Since the counts for Mg, H, and O are the same on both sides, the equation is balanced right from the start.
This step ensures that chemical reactions reflect the true conversions happening at an atomic level, which is vital for any chemist or students learning about chemical reactions.
Metal Hydroxide Formation
Metal hydroxides, like magnesium hydroxide formed here, are a type of compound that is composed of metal cations and hydroxide anions (\(\text{OH}^-\)). In the case of the magnesium oxide and water reaction, magnesium cations from \(\text{MgO}\) join with hydroxide anions that originate from water.
The process of forming a metal hydroxide involves the following:
  • When magnesium oxide (\(\text{MgO}\)) meets water, it starts to dissociate.
  • Magnesium ions (\(\text{Mg}^{2+}\)) unite with hydroxide ions (\(\text{OH}^-\))
  • This forms the compound magnesium hydroxide (\(\text{Mg(OH)}_2\))
Metal hydroxides generally exhibit basic properties. Magnesium hydroxide, for instance, is slightly soluble in water, mean it forms a suspension with water. In its solid form, it is often used for relief from acid indigestion, demonstrating how the formation of these compounds is not just a theoretical exercise, but one with everyday significance.

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Most popular questions from this chapter

Balance the following equations: (a) for the synthesis of urea, a common fertilizer \(\mathrm{CO}_{2}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (b) for the reactions used to make uranium(VI) fluoride for the enrichment of natural uranium \(\mathrm{UO}_{2}(\mathrm{s})+\mathrm{HF}(\mathrm{aq}) \rightarrow \mathrm{UF}_{4}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\) \(\mathrm{UF}_{4}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{g}) \rightarrow \mathrm{UF}_{6}(\mathrm{s})\) (c) for the reaction to make titanium(IV) chloride, which is then converted to titanium metal \(\mathrm{TiO}_{2}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g})+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{TiCl}_{4}(\ell)+\mathrm{CO}(\mathrm{g})\) \(\operatorname{TiCl}_{4}(\ell)+\operatorname{Mg}(\mathrm{s}) \rightarrow \operatorname{Ti}(\mathrm{s})+\operatorname{Mg} \mathrm{Cl}_{2}(\mathrm{s})\)

Describe how to prepare zinc chloride by (a) an acid-base reaction, (b) a gas- forming reaction, and (c) an oxidation-reduction reaction. The available starting materials are \(\mathrm{ZnCO}_{3}, \mathrm{HCl}, \mathrm{Cl}_{2}, \mathrm{HNO}_{3}\) \(\mathrm{Zn}(\mathrm{OH})_{2}, \mathrm{NaCl}, \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2},\) and \(\mathrm{Zn} .\) Write complete balanced equations for the reactions chosen.

You are given mixtures containing the following compounds. Which compound in each pair could be separated by stirring the solid mixture with water? (a) \(\mathrm{NaOH}\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\) (b) \(\mathrm{MgCl}_{2}\) and \(\mathrm{MgF}_{2}\) (c) AgI and KI (d) \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{PbCl}_{2}\)

The following compounds are water-soluble. What ions are produced by each compound in aqueous solution? (a) \(\mathrm{KOH}\) (c) \(\mathrm{LiNO}_{3}\) (b) \(\mathrm{K}_{2} \mathrm{SO}_{4}\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)

Heating \(\mathrm{HI}(\mathrm{g})\) at \(425^{\circ} \mathrm{C}\) causes some of this compound to decompose, forming \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{I}_{2}(\mathrm{g}) .\) Eventually, the amounts of the three species do not change further; the system has reached equilibrium. (At this point, approximately \(22 \%\) of the HI has decomposed.) Describe what is happening in this system at the molecular level.

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