Complete and balance the following: (a) \(\operatorname{As}(s)+\) excess \(\mathrm{O}_{2}(g) \longrightarrow\) (b) \(\mathrm{Bi}(s)+\operatorname{excess} \mathrm{F}_{2}(g) \longrightarrow\) (c) \(\mathrm{Ca}_{3} \mathrm{As}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\)

Short Answer

Expert verified
(a) 4As(s) + 3O2(g) -> 2As2O3(s), (b) 2Bi(s) + 3F2(g) -> 2BiF3(s), (c) Ca3As2(s) + 6H2O(l) -> 3Ca(OH)2(aq) + 2AsH3(g)

Step by step solution

01

- Identify the products for each reaction

Excess oxygen will react with arsenic to form arsenic oxide. Excess fluorine will react with bismuth to form bismuth fluoride. Calcium arsenide reacts with water to produce calcium hydroxide and arsine gas.
02

- Write the unbalanced equations for each reaction

For arsenic and oxygen: \(\text{As}(s) + \text{O}_2(g) \rightarrow \text{As}_2\text{O}_3(s)\)For bismuth and fluorine: \(\text{Bi}(s) + \text{F}_2(g) \rightarrow \text{BiF}_3(s)\)For calcium arsenide and water: \[\text{Ca}_3\text{As}_2(s) + \text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + \text{AsH}_3(g)\]
03

- Balance each chemical equation

Balance the arsenic oxide reaction: \(4\text{As}(s) + 3\text{O}_2(g) \rightarrow 2\text{As}_2\text{O}_3(s)\)Balance the bismuth fluoride reaction: \(2\text{Bi}(s) + 3\text{F}_2(g) \rightarrow 2\text{BiF}_3(s)\)Balance the calcium arsenide reaction: \[\text{Ca}_3\text{As}_2(s) + 6\text{H}_2\text{O}(l) \rightarrow 3\text{Ca(OH)}_2(aq) + 2\text{AsH}_3(g)\]

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Key Concepts

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

Oxidation Reactions
Oxidation reactions are a type of chemical reaction where a substance loses electrons. This process either involves the addition of oxygen or the removal of hydrogen from a compound. For instance, in the reaction between arsenic and oxygen forming arsenic oxide, arsenic loses electrons to oxygen, resulting in its oxidation.
  • Oxidation involves an increase in the oxidation state of the substance.
  • The substance gaining electrons undergoes reduction.
In the oxidation of arsenic, the oxygen acts as the oxidizing agent because it accepts the electrons released by arsenic. Contrastingly, in a reduction reaction, a substance gains electrons and decreases its oxidation state. Understanding oxidation is crucial in balancing chemical equations like the ones provided in the exercise.
Stoichiometry
Stoichiometry is the area of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It is based on the principle of the conservation of mass, meaning that the total mass of reactants in a chemical reaction equals the total mass of the products. When balancing chemical equations, such as the reactions in the exercise, stoichiometry helps ensure that the number of atoms for each element is equal on both sides of the equation.
  • It helps predict the amounts of substances consumed and produced.
  • Coefficients are used to balance the equations.
For example, in the equation for bismuth and fluorine, to balance it using stoichiometry, you adjust the coefficients of the reactants and products to ensure that there are equal numbers of each type of atom on both sides of the reaction.
Chemical Reactions
Chemical reactions occur when substances interact to form new substances with different properties. They are often categorized by the type of reactants and products involved, and the type of change that occurs. The three reactions given in the exercise exemplify different kinds of chemical reactions: the formation of oxides and fluorides, and the reaction of a compound with water to produce hydroxides and gases.
  • A chemical equation represents a chemical reaction.
  • Reactants are the starting substances, and products are the substances formed.
For instance, in the calcium arsenide and water reaction, the reactants (calcium arsenide and water) react to form the products (calcium hydroxide and arsine gas), showcasing how chemical reactions lead to the formation of new compounds from the original substances.

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Most popular questions from this chapter

Chlorine is used to make bleach solutions containing \(5.25 \%\) NaClO (by mass). Assuming \(100 \%\) yield in the reaction producing \(\mathrm{NaClO}\) from \(\mathrm{Cl}_{2}\), how many liters of \(\mathrm{Cl}_{2}(g)\) at STP will be needed to make \(1000 .\) L of bleach solution \((d=1.07 \mathrm{~g} / \mathrm{mL}) ?\)

Rank the following in order of increasing electrical conductivity, and explain your ranking: \(\mathrm{Po}, \mathrm{S},\) Se.

Gaseous \(\mathrm{F}_{2}\) reacts with water to form \(\mathrm{HF}\) and \(\mathrm{O}_{2}\). In \(\mathrm{NaOH}\) solution, \(\mathrm{F}_{2}\) forms \(\mathrm{F}^{-}\), water, and oxygen difluoride (OF \(_{2}\) ), a highly toxic gas and powerful oxidizing agent. The OF \(_{2}\) reacts with excess \(\mathrm{OH}^{-}\), forming \(\mathrm{O}_{2}\), water, and \(\mathrm{F}\) (a) For each reaction, write a balanced cquation, give the oxidation state of \(\mathrm{O}\) in all compounds, and identify the oxidizing and reducing agents. (b) Draw a Lewis structure for \(\mathrm{OF}_{2}\), and predict its shape.

Give the name and symbol or formula of a Group \(6 \mathrm{~A}(16)\) element or compound that fits each description or use: (a) Unstable allotrope of oxygen (b) Oxide having sulfur with the same O.N. as in sulfuric acid (c) Air pollutant produced by burning sulfur-containing coal (d) Powerful dehydrating agent (e) Compound used in solution in the photographic process

From its formula, one might expect \(\mathrm{CO}\) to be quite polar, but its dipole moment is actually small \((0.11 \mathrm{D})\). (a) Draw the Lewis structure for CO. (b) Calculate the formal charges. (c) Based on your answers to parts (a) and (b), explain why the dipole moment is so small.

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