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Chapter 7: Problem 78

Write a balanced equation for the reaction that occurs in each (a) Calcium metal is heated in an atmoof the following cases: sphere of oxygen gas. (b) Copper oxide is heated in an atmosphere of hydrogen gas. (c) Chlorine reacts with nitrogen gas. (d) Boron tribromide reacts with water.

Short Answer

Expert verified
The balanced equations for the given reactions are: (a) 2Ca + O2 → 2CaO (b) CuO + H2 → Cu + H2O (c) 3Cl2 + N2 → 2NCl3 (d) BBr3 + 3H2O → H3BO3 + 3HBr

Step by step solution

01

(a) Identifying the reaction between calcium and oxygen

To determine the product of the reaction between calcium metal (Ca) and oxygen gas (O2), we need to know the valency of the calcium and oxygen. Calcium has a valency of +2 and oxygen has a valency of -2. Based on the valency, calcium and oxygen come together to form calcium oxide (CaO). Now we need to balance the equation: Ca + O2 → CaO.
02

(a) Balancing the equation

To balance the equation, we need to make sure that the number of atoms of each element is equal on both sides. Ca + O2 → CaO 1Ca + 2O → 1Ca + 1O Since there are 2 oxygen atoms on the left side but only 1 on the right side, multiply the right side by 2 to balance the equation: Ca + O2 → 2CaO Now, we need to balance the number of calcium atoms. Multiply the calcium atom on the left side of the equation by 2: 2Ca + O2 → 2CaO Now the equation is balanced.
03

(b) Identifying the reaction between copper oxide and hydrogen

To determine the product of the reaction between copper(II) oxide (CuO) and hydrogen gas (H2), we need to know that hydrogen is a reducing agent that will reduce CuO to elemental copper (Cu) and form water (H2O) as a byproduct: CuO + H2 → Cu + H2O.
04

(b) Balancing the equation

To balance the equation, we need to make sure that the number of atoms of each element is equal on both sides. CuO + H2 → Cu + H2O 1Cu + 1O + 2H → 1Cu + 2H + 1O The equation is already balanced.
05

(c) Identifying the reaction between chlorine and nitrogen

The reaction between chlorine gas (Cl2) and nitrogen gas (N2) produces nitrogen trichloride (NCl3). Now we need to balance the equation: Cl2 + N2 → NCl3.
06

(c) Balancing the equation

To balance the equation, we need to make sure the number of atoms of each element is equal on both sides. Cl2 + N2 → NCl3 2Cl + 2N → 1N + 3Cl To balance the number of nitrogen atoms, multiply the NCl3 on the right side by 2: Cl2 + N2 → 2NCl3 2Cl + 2N → 2N + 6Cl To balance the number of chlorine atoms, multiply the Cl2 on the left side by 3: 3Cl2 + N2 → 2NCl3 6Cl + 2N → 2N + 6Cl Now the equation is balanced.
07

(d) Identifying the reaction between boron tribromide and water

The reaction between boron tribromide (BBr3) and water (H2O) produces boric acid (H3BO3) and hydrobromic acid (HBr). Now we need to balance the equation: BBr3 + H2O → H3BO3 + HBr.
08

(d) Balancing the equation

To balance the equation, we need to make sure the number of atoms of each element is equal on both sides. BBr3 + H2O → H3BO3 + HBr 1B + 3Br + 2H + 1O → 1B + 1O + 3H + 1H + 1Br To balance the number of bromine atoms, multiply HBr on the right side by 3: BBr3 + H2O → H3BO3 + 3HBr 1B + 3Br + 2H + 1O → 1B + 1O + 3H + 3H + 3Br Now the equation is balanced.

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Key Concepts

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

Balanced Equations
A chemical equation is balanced when the number of atoms of each element is equal on both sides of the equation. This is crucial because it follows the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction.
To balance an equation, you can follow these simple steps:
  • Write down the unbalanced equation.
  • Count the number of atoms of each element on both sides.
  • Add coefficients (not subscripts) in front of compounds to balance the atoms.
  • Check your work to ensure all atoms balance out.
This ensures that your reaction reflects what actually happens during the chemical process. Always remember: never change the chemical formula of a compound to balance an equation.
Calcium and Oxygen Reaction
When calcium (Ca) reacts with oxygen (Oâ‚‚), they form calcium oxide (CaO). Calcium has a valency of +2, meaning it tends to lose two electrons, while oxygen has a valency of -2, meaning it tends to gain two electrons.
This makes calcium and oxygen a perfect pair, creating a stable ionic compound, CaO.
To balance the equation:
  • Start with Ca + Oâ‚‚ → CaO.
  • Notice the imbalance in oxygen atoms: 2 on the left, 1 on the right.
  • Place a coefficient of 2 in front of CaO to get Ca + Oâ‚‚ → 2CaO.
  • Balance calcium by placing a 2 in front of Ca: 2Ca + Oâ‚‚ → 2CaO.
Now, both sides have equal numbers of each atom, making the equation balanced.
Copper Oxide and Hydrogen Reaction
Copper(II) oxide (CuO) reacts with hydrogen gas (Hâ‚‚) to produce copper (Cu) and water (Hâ‚‚O). In this reaction, hydrogen acts as a reducing agent, meaning it helps reduce copper oxide to copper.
The initial equation is CuO + H₂ → Cu + H₂O.
Look at the equation:
  • It is already balanced because each side contains 1 Cu, 1 O, and 2 H atoms.
In such reactions, hydrogen's role in reducing metals is essential in various industrial applications.
Chlorine and Nitrogen Reaction
Chlorine gas (Cl₂) and nitrogen gas (N₂) react to form nitrogen trichloride (NCl₃). This process produces a compound often used in the industrial sector.
To balance the initial equation, Cl₂ + N₂ → NCl₃:
  • The unbalanced version shows 2 Cl from Clâ‚‚ and 2 N from Nâ‚‚ reacting to form 1 N and 3 Cl in NCl₃.
  • To balance nitrogen, place a coefficient of 2 before NCl₃: Clâ‚‚ + Nâ‚‚ → 2NCl₃.
  • Now, balance chlorine by placing a 3 before Clâ‚‚: 3Clâ‚‚ + Nâ‚‚ → 2NCl₃.
This balanced equation now accurately represents the reaction.

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Most popular questions from this chapter

Which will experience the greater effect nuclear charge, the electrons in the \(n=2\) shell in \(\mathrm{F}\) or the \(n=2\) shell in \(\mathrm{B}\) ? Which will be closer to the nucleus?

Elements in group 17 in the periodic table are called the halogens; elements in group 16 are called the chalcogens. (a) What is the most common oxidation state of the chalcogens compared to the halogens? (b) For each of the following periodic properties, state whether the halogens or the chalcogens have larger values: atomic radii, ionic radii of the most common oxidation state, first ionization energy, second ionization energy.

(a) Does metallic character increase, decrease, or remain unchanged as one goes from left to right across a row of the periodic table? (b) Does metallic character increase, decrease, or remain unchanged as one goes down a column of the periodic table? (c) Are the periodic trends in (a) and (b) the same as or different from those for first ionization energy?

Hydrogen is an unusual element because it behaves in some ways like the alkali metal elements and in other ways like nonmetals. Its properties can be explained in part by its electron configuration and by the values for its ionization energy and electron affinity. (a) Explain why the electron affinity of hydrogen is much closer to the values for the alkali elements than for the halogens. (b) Is the following statement true? "Hydrogen has the smallest bonding atomic radius of any element that forms chemical compounds." If not, correct it. If it is, explain in terms of electron configurations. (c) Explain why the ionization energy of hydrogen is closer to the values for the halogens than for the alkali metals. (d) The hydride ion is \(\mathrm{H}^{-}\). Write out the process corresponding to the first ionization energy of the hydride ion. (e) How does the process in part (d) compare to the process for the electron affinity of a neutral hydrogen atom?

Write balanced equations for the following reactions: (a) sulfur dioxide with water, (b) lithium oxide in water, \((\mathbf{c})\) zinc oxide with dilute hydrochloric acid, (d) arsenic trioxide with aqueous potassium hydroxide.
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