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. ${\mathrm{C}}_2{\mathrm{H}}_2$
5. CO

Short answer

(a) Balanced chemical equation of Mg is $2\mathrm{M}\mathrm{g}(\mathrm{s})+{\mathrm{O}}_2(\mathrm{g})\to 2\mathrm{M}\mathrm{g}\mathrm{O}(\mathrm{s})$

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

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

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

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

Step-by-step solution

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.

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:

$\mathrm{M}\mathrm{g}(\mathrm{s})+{\mathrm{O}}_2(\mathrm{g})\to \mathrm{M}\mathrm{g}\mathrm{O}(\mathrm{s})$

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:

$2\mathrm{M}\mathrm{g}(\mathrm{s})+{\mathrm{O}}_2(\mathrm{g})\to 2\mathrm{M}\mathrm{g}\mathrm{O}(\mathrm{s})$

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:

$\mathrm{R}\mathrm{b}(\mathrm{s})+{\mathrm{O}}_2(\mathrm{g})\to \mathrm{R}{\mathrm{b}}_2\mathrm{O}(\mathrm{s})$

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 $\mathrm{R}{\mathrm{b}}_2\mathrm{O}$by 2 to get the following:

$4\mathrm{R}\mathrm{b}(\mathrm{s})+{\mathrm{O}}_2(\mathrm{g})\to 2\mathrm{R}{\mathrm{b}}_2\mathrm{O}(\mathrm{s})$

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.

$\mathrm{G}\mathrm{a}(\mathrm{s})+{\mathrm{O}}_2(\mathrm{g})\to \mathrm{G}{\mathrm{a}}_2{\mathrm{O}}_3(\mathrm{s})$

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, ${\mathrm{O}}_2$ by 3, and $\mathrm{G}{\mathrm{a}}_2{\mathrm{O}}_3$by 2, as follows:

$4\mathrm{G}\mathrm{a}(\mathrm{s})+3{\mathrm{O}}_2(\mathrm{g})\to 2\mathrm{G}{\mathrm{a}}_2{\mathrm{O}}_3(\mathrm{s})$

Find the balanced chemical equation of ${\mathrm{C}}_2{\mathrm{H}}_2$

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

${\mathrm{C}}_2{\mathrm{H}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})\to \mathrm{C}{\mathrm{O}}_2(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

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

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

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:

$\mathrm{C}\mathrm{O}(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})\to \mathrm{C}{\mathrm{O}}_2(\mathrm{g})$

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 $\mathrm{C}{\mathrm{O}}_2$by 2 to get:

$2\mathrm{C}\mathrm{O}(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})\to 2\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$