Question

Balance the following equations: (a) \(\mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) (b) \(\mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HNO}_{3}(a q)\) (c) \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(l)+\operatorname{HCl}(g)\) (d) \(\mathrm{Zn}(\mathrm{OH})_{2}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

The balanced equations are: (a) \(CO(g) + O_2(g) \rightarrow CO_2(g)\) (b) \(N_2O_5(g) + 2H_2O(l) \rightarrow 2HNO_3(aq)\) (c) \(CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(l) + 4HCl(g)\) (d) \(Zn(OH)_2(s) + 2HNO_3(aq) \rightarrow Zn(NO_3)_2(aq) + 2H_2O(l)\)

Step-by-step solution

(a) Balance CO(g) + O2(g) → CO2(g)

To balance this equation, we need to make sure that the number of carbon and oxygen atoms are the same on both sides. In this case, the equation is already balanced because there is one carbon and two oxygen atoms on each side. Therefore, the balanced equation is: \(CO(g) + O_2(g) \rightarrow CO_2(g)\)

(b) Balance N2O5(g) + H2O(l) → HNO3(aq)

First, we need to balance the nitrogen atoms. The left side has two nitrogen atoms and the right side has one nitrogen atom. We can balance the nitrogen atoms by placing a 2 in front of HNO3: \(N_2O_5(g) + H_2O(l) \rightarrow 2HNO_3(aq)\) Next, we need to balance the oxygen atoms. The left side has five oxygen atoms and the right side has six oxygen atoms. The equation is already balanced for oxygen, as there are five oxygen atoms in N2O5 and one oxygen atom in H2O, totaling six oxygen atoms. Finally, we need to balance the hydrogen atoms. The left side has two hydrogen atoms and the right side has four hydrogen atoms. We can balance the hydrogen atoms by placing a 2 in front of H2O: \(N_2O_5(g) + 2H_2O(l) \rightarrow 2HNO_3(aq)\)

(c) Balance CH4(g) + Cl2(g) → CCl4(l) + HCl(g)

First, let's balance the carbon atoms. The equation is already balanced for carbon, with one carbon atom on each side. Next, let's balance the hydrogen atoms. The left side has four hydrogen atoms and the right side has one hydrogen atom. We can balance the hydrogen atoms by placing a 4 in front of HCl: \(CH_4(g) + Cl_2(g) \rightarrow CCl_4(l) + 4HCl(g)\) Finally, let's balance the chlorine atoms. The left side has two chlorine atoms and the right side has eight chlorine atoms. We can balance the chlorine atoms by placing a 4 in front of Cl2: \(CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(l) + 4HCl(g)\)

(d) Balance Zn(OH)2(s) + HNO3(aq) → Zn(NO3)2(aq) + H2O(l)

First, let's balance the zinc atoms. The equation is already balanced for zinc, with one zinc atom on each side. Next, let's balance the nitrogen atoms. The left side has one nitrogen atom and the right side has two nitrogen atoms. We can balance the nitrogen atoms by placing a 2 in front of HNO3: \(Zn(OH)_2(s) + 2HNO_3(aq) \rightarrow Zn(NO_3)_2(aq) + H_2O(l)\) Now, let's balance the oxygen atoms. The left side has five oxygen atoms (two oxygen atoms in the hydroxide and three oxygen atoms in each nitrate ion) and the right side has eight oxygen atoms. To balance the oxygen atoms, the equation is already balanced, as there are two oxygen atoms in Zn(OH)2 and six oxygen atoms in 2HNO3, totaling eight oxygen atoms. Finally, let's balance the hydrogen atoms. The left side has four hydrogen atoms (two hydrogen atoms in the hydroxide and two hydrogen atoms in each nitrate ion) and the right side has two hydrogen atoms. To balance the hydrogen atoms, we need to place a 2 in front of H2O: \(Zn(OH)_2(s) + 2HNO_3(aq) \rightarrow Zn(NO_3)_2(aq) + 2H_2O(l)\)

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. It is based on the law of conservation of mass, which states that matter cannot be created or destroyed. In simpler terms, stoichiometry allows us to understand the ratios of molecules involved in reactions.
For example, when given a chemical equation, stoichiometry helps determine how much of each substance is needed to completely react with another. This is crucial when performing laboratory experiments or when industrially producing chemicals, as it ensures efficiency and cost-effectiveness.

• Understanding stoichiometry begins with a balanced chemical equation.
• Coefficients in the balanced equation indicate the ratio of moles of each substance involved.
• Using these coefficients, we can calculate the mass and volume of reactants and products. For example, in the equation \(CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O\), two moles of \(O_2\) are needed for every mole of \(CH_4\).

This application of mole ratios aids in predicting product yields and necessary reactant amounts.

Reaction Coefficients

Reaction coefficients are essential in balancing chemical equations. They tell us how many molecules or moles of a substance participate in a reaction. Balancing equations ensures the number of each type of atom is equal on both sides of the reaction.
The process of using reaction coefficients involves:

• Writing down the unbalanced equation.
• Identifying the number of atoms for each element on both the reactant and product sides.
• Adding coefficients to molecules to equalize the number of each atom on both sides.

For example, in the unbalanced reaction of \(N_2O_5 + H_2O \rightarrow HNO_3\), coefficients are applied to balance the nitrogen, oxygen, and hydrogen atoms. Coefficients play a crucial role in stoichiometry as they dictate the mole ratio of reactants to products, allowing for accurate calculations in chemical processes.

Oxygen Balancing

Oxygen balancing is a specific step needed when balancing chemical equations. Oxygen is often present in multiple compounds within a reaction, requiring careful adjustment of coefficients to balance.
When balancing oxygen, follow these guidelines:

• Count the oxygen atoms on both reactant and product sides.
• Add coefficients to compounds containing oxygen to equalize the atom count across the equation.
• Recheck the entire equation after balancing oxygen to ensure that no other elements are left unbalanced.

For example, in the equation \(CH_4 + 4Cl_2 \rightarrow CCl_4 + 4HCl\), chlorine and hydrogen balancing must be achieved first before confirming oxygen balance. Ensuring oxygen is balanced is vital as it typically appears in various crucial compounds like \(H_2O\) and \(CO_2\). Proper oxygen balancing affects the accuracy of most stoichiometric calculations.

Hydrogen Balancing

Balancing hydrogen in a chemical equation is crucial because many reactions involve water or other hydrogen-containing compounds. Proper hydrogen balancing helps in maintaining the principle of mass conservation.
Steps for hydrogen balancing include:

• Counting the hydrogen atoms in all reacting substances and products.
• Adjusting coefficients to equalize hydrogen atoms on both sides.
• Checking that other elements remain balanced afterward.

For instance, in \(Zn(OH)_2 + 2HNO_3 \rightarrow Zn(NO_3)_2 + 2H_2O\), hydrogen balancing is needed, especially when water is a product. The hydrogen atoms in \(Zn(OH)_2\) and \(HNO_3\) must match those in the resulting \(H_2O\). Accurately balancing hydrogen atoms ensures that the overall reaction maintains the integrity of the chemical identity.