Problem 67 - Chemistry: The Central Science in SI Units, Global Edition

Question

Write a balanced equation for each of the following reactions: (a) Hydrogen cyanide is formed commercially by passing a mixture of methane, ammonia, and air over a catalyst at \(800^{\circ} \mathrm{C}\). Water is a by-product of the reaction. (b) Baking soda reacts with acids to produce carbon dioxide gas. (c) When barium carbonate reacts in air with sulfur dioxide, barium sulfate and carbon dioxide form.

Step-by-Step Solution

Answer

The balanced equations for the given reactions are: a) \(2\mathrm{CH}_4 + 2\mathrm{NH}_3 + 3\mathrm{O}_2 \rightarrow 2\mathrm{HCN} + 6\mathrm{H}_2\mathrm{O}\) b) \(\mathrm{NaHCO}_3 + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2\) c) \(\mathrm{BaCO}_3 + \mathrm{SO}_2 \rightarrow \mathrm{BaSO}_4 + \mathrm{CO}_2\)

Step 1: Identify reactants and products

For the formation of hydrogen cyanide (HCN), the reactants given are methane (CH4), ammonia (NH3), and air (which mainly consists of O2). The products mentioned are hydrogen cyanide and water (H2O).

Step 2: Write an unbalanced equation

Write down the unbalanced equation by placing the reactants on the left side and products on the right side of the equation: CH4 + NH3 + O2 -> HCN + H2O

Step 3: Balance the equation

Balance the number of atoms for each element on both sides of the equation: 2CH4 + 2NH3 + 3O2 -> 2HCN + 6H2O The balanced equation for the formation of hydrogen cyanide is: \(2\mathrm{CH}_4 + 2\mathrm{NH}_3 + 3\mathrm{O}_2 \rightarrow 2\mathrm{HCN} + 6\mathrm{H}_2\mathrm{O}\) b) Baking soda and acids producing carbon dioxide

Step 1: Identify reactants and products

When baking soda (NaHCO3) reacts with an acid, such as hydrochloric acid (HCl), the products formed are sodium chloride (NaCl), water (H2O), and carbon dioxide (CO2).

Step 2: Write an unbalanced equation

Write the unbalanced equation: NaHCO3 + HCl -> NaCl + H2O + CO2

Step 3: Balance the equation

This equation is already balanced with one mole of each substance: \( \mathrm{NaHCO}_3 + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2\) c) Barium carbonate and sulfur dioxide reaction

Step 1: Identify reactants and products

In this reaction, the reactants are barium carbonate (BaCO3) and sulfur dioxide (SO2). The products formed are barium sulfate (BaSO4) and carbon dioxide (CO2).

Step 2: Write an unbalanced equation

Write the unbalanced equation: BaCO3 + SO2 -> BaSO4 + CO2

Step 3: Balance the equation

This equation is also already balanced with one mole of each substance: \( \mathrm{BaCO}_3 + \mathrm{SO}_2 \rightarrow \mathrm{BaSO}_4 + \mathrm{CO}_2\)

Key Concepts

Balanced Equations

Balanced equations are a fundamental part of understanding chemical reactions. They ensure that there are equal numbers of each type of atom on both sides of the equation, maintaining the law of conservation of mass. This is crucial because atoms are neither created nor destroyed in chemical reactions.

When balancing an equation, start by writing down the unbalanced equation with all the reactants on the left and products on the right. For example, in the formation of hydrogen cyanide (HCN) from methane (CH extsubscript{4}), ammonia (NH extsubscript{3}), and oxygen ( ext{O} extsubscript{2}), you would first write the unbalanced equation:

• CH extsubscript{4} + NH extsubscript{3} + O extsubscript{2} → HCN + H extsubscript{2}O

Next, adjust coefficients to balance each element in the reaction. This generally means starting with the most complex molecule and working towards the simplest. The balanced equation here becomes:

\[2 ext{CH}_4 + 2 ext{NH}_3 + 3 ext{O}_2 ightarrow 2 ext{HCN} + 6 ext{H}_2 ext{O}\] By ensuring that each atom's count remains constant throughout the reaction, we confirm that our equation is balanced.

Stoichiometry

Stoichiometry is a section of chemistry that involves using ratios derived from balanced equations to calculate amounts of reactants or products. It's like a recipe in cooking, where specific amounts of ingredients yield a certain result. This helps understand the quantitative relationships in chemical reactions.

In our example of baking soda reacting with hydrochloric acid, stoichiometry helps us understand how much product we can expect from a given amount of reactant. For the balanced equation:

• NaHCO extsubscript{3} + HCl → NaCl + H extsubscript{2}O + CO extsubscript{2}

We see that one mole of NaHCO extsubscript{3} reacts with one mole of HCl to produce one mole of NaCl, one mole of water, and one mole of carbon dioxide gas. This kind of quantitative understanding enables chemists to plan and predict the outcomes of experiments with precision.

Such calculations are essential for industries that rely on specific product yields, such as pharmaceuticals, agriculture, and materials manufacturing.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show how reactants transform into products, using chemical symbols and formulae. This provides a concise way to convey complex chemical concepts.

An essential part of chemical equations is understanding the state symbols that denote whether a substance is solid (s), liquid (l), gas (g), or aqueous (aq). These symbols offer insights into the conditions under which the reaction takes place.

In the reaction of barium carbonate and sulfur dioxide:

• BaCO extsubscript{3} + SO extsubscript{2} → BaSO extsubscript{4} + CO extsubscript{2}

Each reactant and product is expressed in a simple yet informative manner. The balanced equation not only tells us about the reactants and products but also gives hints about the reaction conditions if additional information is provided, like catalysts or temperatures. Chemical equations are foundational to studying how chemicals interact and transform, leading to myriad practical applications across scientific and industrial fields.