Q88 E - Chemistry (OpenStax)

Question

Write a balanced chemical equation for the reaction of an excess of oxygen with each of the following.

 

Remember that oxygen is a strong oxidizing agent and tends to oxidize an element to its maximum oxidation state.

1. Mg.
2. Rb.
3. Ga.
4. \({{\rm{C}}_2}{{\rm{H}}_2}\)
5. CO

Step-by-Step Solution

Answer

(a) Balanced chemical equation of Mg is  \(2{\rm{Mg}}({\rm{s}}) + {{\rm{O}}_2}(\;{\rm{g}}) \to 2{\rm{MgO}}({\rm{s}})\)

(b) Balanced chemical equation of Rb is \(4{\rm{Rb}}({\rm{s}}) + {{\rm{O}}_2}(\;{\rm{g}}) \to 2{\rm{R}}{{\rm{b}}_2}{\rm{O}}({\rm{s}})\)

(c) Balanced chemical equation of Ga is \(2{{\rm{C}}_2}{{\rm{H}}_2}(\;{\rm{g}}) + 5{{\rm{O}}_2}(\;{\rm{g}}) \to 4{\rm{C}}{{\rm{O}}_2}(\;{\rm{g}}) + 2{{\rm{H}}_2}{\rm{O}}({\rm{I}})\)

(d) Balanced chemical equation of \({{\rm{C}}_2}{{\rm{H}}_2}\)is \(2{{\rm{C}}_2}{{\rm{H}}_2}(\;{\rm{g}}) + 5{{\rm{O}}_2}(\;{\rm{g}}) \to 4{\rm{C}}{{\rm{O}}_2}(\;{\rm{g}}) + 2{{\rm{H}}_2}{\rm{O}}({\rm{I}})\)

(e) Balanced chemical equation of CO is \(2{\rm{CO}}({\rm{g}}) + {{\rm{O}}_2}(\;{\rm{g}}) \to 2{\rm{C}}{{\rm{O}}_2}(\;{\rm{g}})\)

 

Step 1: Defintion of acid-base reactions

• In terms of reaction stoichiometry, a balanced equation depicts the amount of reactants consumed or the amount of products produced.
• Stoichiometry is a mathematical formula that calculates the coefficients of the reactants and products in a chemical process. It denotes the mole ratio at which reactants and products mix to generate products.
• Metals react with oxygen to make metal oxides, whereas hydrocarbons normally react with oxygen to produce carbon dioxide and water.

Step 2: Find the balanced chemical equation of Mg

Magnesium forms magnesium oxide when it combines with oxygen. The following is a diagram of the chemical reaction:

 

\({\rm{Mg}}({\rm{s}}) + {{\rm{O}}_2}(\;{\rm{g}}) \to {\rm{MgO}}({\rm{s}})\)

 

Because the number of O atoms on the reactants is 2, but only one on the product side, the reaction is unbalanced. To make it even, multiply Mg and MgO by 2 to get:

 

\(2{\rm{Mg}}({\rm{s}}) + {{\rm{O}}_2}(\;{\rm{g}}) \to 2{\rm{MgO}}({\rm{s}})\)

 

Step 3: Find the balanced chemical equation of Rb

Rubidium oxide is formed when rubidium interacts with oxygen. The following is a diagram of the chemical reaction:

 

\({\rm{Rb}}({\rm{s}}) + {{\rm{O}}_2}(\;{\rm{g}}) \to {\rm{R}}{{\rm{b}}_2}{\rm{O}}({\rm{s}})\)

 

The reaction is unbalanced because the number of Rb and Oatoms on the reactants is 1 and 2, respectively, whereas the number of Rb and Oatoms on the product side is 2 and 1. To balance it, multiply Rb by 4 and \({\rm{R}}{{\rm{b}}_2}{\rm{O}}\)by 2 to get the following: 

 

\(4{\rm{Rb}}({\rm{s}}) + {{\rm{O}}_2}(\;{\rm{g}}) \to 2{\rm{R}}{{\rm{b}}_2}{\rm{O}}({\rm{s}})\)

Step 4: Find the balanced chemical equation of Ga

Gallium oxide is formed when gallium interacts with oxygen. The chemical reaction is depicted in the diagram below.

 

\({\rm{Ga}}({\rm{s}}) + {{\rm{O}}_2}(\;{\rm{g}}) \to {\rm{G}}{{\rm{a}}_2}{{\rm{O}}_3}(\;{\rm{s}})\)

 

The reaction is unbalanced because the number of Ga and O atoms on the reactants is 1 and 2, respectively, while the number of Ga and O atoms on the product side is 2 and 3. To balance it, multiply Ga by 4, \({{\rm{O}}_2}\) by 3, and \({\rm{G}}{{\rm{a}}_2}{{\rm{O}}_3}\)by 2, as follows:

 

\(4{\rm{Ga}}({\rm{s}}) + 3{{\rm{O}}_2}(\;{\rm{g}}) \to 2{\rm{G}}{{\rm{a}}_2}{{\rm{O}}_3}(\;{\rm{s}})\)

Step 5: Find the balanced chemical equation of \({{\rm{C}}_2}{{\rm{H}}_2}\)

When acetylene comes into contact with oxygen, it produces carbon dioxide and water. The following is a diagram of the chemical reaction:

 

\({{\rm{C}}_2}{{\rm{H}}_2}(\;{\rm{g}}) + {{\rm{O}}_2}(\;{\rm{g}}) \to {\rm{C}}{{\rm{O}}_2}(\;{\rm{g}}) + {{\rm{H}}_2}{\rm{O (l) }}\)

 

Because the number of C, H, and O atoms on the reactants and products are unequal, the equation is unbalanced. To balance it, multiply \({{\rm{C}}_2}{{\rm{H}}_2}\)and \({{\rm{O}}_2}\) by 2 and 5 on the reactants side, and \({\rm{C}}{{\rm{O}}_2}\)and \({{\rm{H}}_2}{\rm{O}}\)by 4 and 2 on the products side, respectively, to get:

 

\(2{{\rm{C}}_2}{{\rm{H}}_2}(\;{\rm{g}}) + 5{{\rm{O}}_2}(\;{\rm{g}}) \to 4{\rm{C}}{{\rm{O}}_2}(\;{\rm{g}}) + 2{{\rm{H}}_2}{\rm{O}}({\rm{I}})\)

Step 6: Find the balanced chemical equation of CO

Carbon monoxide is formed when carbon monoxide combines with oxygen. The chemical reaction is depicted in the diagram below:

 

\({\rm{CO}}({\rm{g}}) + {{\rm{O}}_2}(\;{\rm{g}}) \to {\rm{C}}{{\rm{O}}_2}(\;{\rm{g}})\)

The equation is unbalanced because the number of O atoms in the reactants is 3, but only two in the products. To make it even, multiply CO and \({\rm{C}}{{\rm{O}}_2}\)by 2 to get:

 

\(2{\rm{CO}}({\rm{g}}) + {{\rm{O}}_2}(\;{\rm{g}}) \to 2{\rm{C}}{{\rm{O}}_2}(\;{\rm{g}})\)