Problem 33 - Chemistry: Atoms First

Question

Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas: (a) \(\mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2} \mathrm{S}(g) \rightarrow\) (b) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{S}(g) \longrightarrow\)

Step-by-Step Solution

Answer

Balanced equations: (a) \(\mathrm{Ca(OH)_2(s) + H_2S(g) \rightarrow CaS(s) + H_2O(l)}\); (b) \(\mathrm{Na_2CO_3(aq) + 2H_2S(g) \rightarrow 2NaHCO_3(aq) + H_2O(l)}\).

Step 1: Balancing Reaction (a)

Write down the unbalanced equation for reaction (a) with all the reactants and products. The products for reaction with calcium hydroxide and hydrogen sulfide will be calcium sulfide and water. The unbalanced equation is: \(\mathrm{Ca(OH)_2(s) + H_2S(g) \rightarrow CaS(s) + H_2O(l)}\). To balance it, ensure there are equal numbers of each type of atom on both sides of the equation. The equation is already balanced with one molecule of calcium hydroxide reacting with one molecule of hydrogen sulfide to produce one molecule of calcium sulfide and one molecule of water.

Step 2: Balancing Reaction (b)

Write down the unbalanced equation for reaction (b) with all the reactants and products. The products for the reaction with sodium carbonate and hydrogen sulfide will be sodium bicarbonate and water. The unbalanced equation is: \(\mathrm{Na_2CO_3(aq) + H_2S(g) \rightarrow NaHCO_3(aq) + H_2O(l)}\). To balance it, note that two sodium atoms from sodium carbonate will result in two sodium bicarbonate molecules. The balanced equation is: \(\mathrm{Na_2CO_3(aq) + 2H_2S(g) \rightarrow 2NaHCO_3(aq) + H_2O(l)}\). There are now two sodium atoms, one carbonate ion, two hydrogen sulfides reacting to form two sodium bicarbonates and one water molecule.

Key Concepts

Stoichiometry

Stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It is based on the conservation of mass and the principle that atoms are neither created nor destroyed in chemical reactions. In other words, the amount of each element must be the same on both sides of a balanced chemical equation.

For instance, in the reaction between calcium hydroxide and hydrogen sulfide, the stoichiometry is straightforward. The balanced equation, \(\mathrm{Ca(OH)_2(s) + H_2S(g) \rightarrow CaS(s) + H_2O(l)}\), indicates that one mole of calcium hydroxide reacts with one mole of hydrogen sulfide to produce one mole of calcium sulfide and one mole of water. No coefficients are needed because the number of atoms for each element is equal on both sides of the reaction.

In contrast, reaction (b) is more complex and requires stoichiometric analysis to balance it correctly. The equation \(\mathrm{Na_2CO_3(aq) + H_2S(g) \rightarrow NaHCO_3(aq) + H_2O(l)}\) becomes balanced when we realize that two moles of hydrogen sulfide are required to react with one mole of sodium carbonate. After balancing, the equation reads \(\mathrm{Na_2CO_3(aq) + 2H_2S(g) \rightarrow 2NaHCO_3(aq) + H_2O(l)}\), showing clear stoichiometric proportions.

Understanding stoichiometry is essential for predicting the amounts of products formed and reactants used in chemical reactions, making it a foundational concept in chemistry.

Chemical Reactions

Chemical reactions involve the transformation of one or more substances into new substances. They are described by chemical equations that show the reactants on the left side of an arrow and the products on the right. To make sense of these transformations, chemists use symbols and formulas to represent the different elements and compounds involved.

In the reactions outlined in the exercise, \(\mathrm{Ca(OH)_2}\) and \(\mathrm{H_2S}\) undergo a chemical change to form \(\mathrm{CaS}\) and \(\mathrm{H_2O}\). Meanwhile, in the second reaction, \(\mathrm{Na_2CO_3}\) reacts with \(\mathrm{H_2S}\) to produce \(\mathrm{NaHCO_3}\) and \(\mathrm{H_2O}\). These equations summarize the changes at the molecular level. Balancing these equations ensures that the law of conservation of mass is upheld.

Throughout these processes, different types of chemical reactions can be identified. The first reaction, involving the production of calcium sulfide and water, is a type of double displacement reaction. The second, leading to sodium bicarbonate and water, involves an acid-base reaction with a carbonate acting as the base. Recognizing the type of reaction can help predict the products and ensure equations are balanced correctly.

Hydrogen Sulfide Removal

Hydrogen sulfide (\(\mathrm{H_2S}\)) is an undesirable and toxic component often found in natural gas. Its removal is crucial for safety, environmental, and processing reasons. The reactions provided in the exercise are examples of how hydrogen sulfide can be removed from natural gas streams.

In reaction (a), \(\mathrm{H_2S}\) gas reacts with solid calcium hydroxide \(\mathrm{Ca(OH)_2}\) to form calcium sulfide \(\mathrm{CaS}\) and liquid water \(\mathrm{H_2O}\). This process is known as 'gas sweetening' because it effectively removes the 'sour' \(\mathrm{H_2S}\) from the gas.

Benefits of Calcium Hydroxide

Calcium hydroxide is an economical and easily handled reagent that can neutralize acidic gases like \(\mathrm{H_2S}\).

Reaction (b) demonstrates the use of aqueous sodium carbonate \(\mathrm{Na_2CO_3}\) to convert \(\mathrm{H_2S}\) into sodium bicarbonate \(\mathrm{NaHCO_3}\) and water.

Advantages of Sodium Carbonate

Sodium carbonate has a high capacity for absorbing \(\mathrm{H_2S}\) and can be used in wet scrubbing processes. Both methods effectively remove \(\mathrm{H_2S}\), preventing corrosion and the release of harmful gases into the atmosphere.