Problem 11 - Chemistry The Central Science In Si Units

Question

Balance the following equations: (a) \(\mathrm{SiCl}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Si}(\mathrm{OH})_{4}(s)+\mathrm{HCl}(a q)\) (b) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Step-by-Step Solution

Answer

(a) Balance with 4 \(\mathrm{H}_{2} \mathrm{O}\), (b) Use coefficients 6 for \(\mathrm{CO}_{2}\) and \(\mathrm{O}_{2}\), (c) 2 \(\mathrm{Al}(\mathrm{OH})_{3}\), (d) Use 2 \(\mathrm{H}_{3} \mathrm{PO}_{4}\).

Step 1: Balancing Silicon Reaction

For the reaction \(\mathrm{SiCl}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Si} ( \mathrm{OH} )_{4}(s) +\mathrm{HCl}(aq)\), start by balancing the silicon atoms. Both sides have 1 silicon atom. Next, balance the chlorine atoms; there are 4 chlorine atoms in \(\mathrm{SiCl}_{4}\), so you need 4 \(\mathrm{HCl}\) molecules on the right side. Finally, balance the hydrogen and oxygen atoms by ensuring there are 4 water molecules. The balanced equation is: \[ \mathrm{SiCl}_{4}(l) + 4\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Si}(\mathrm{OH})_{4}(s) + 4\mathrm{HCl}(aq) \]

Step 2: Balancing Glucose Formation Reaction

For the reaction \(\mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) + \mathrm{O}_{2}(g)\), start by balancing the carbon atoms. Place a coefficient of 6 in front of \(\mathrm{CO}_{2}\) to match the 6 carbons in glucose. Next, balance the hydrogen atoms by adding 6 water molecules. Balance the oxygen atoms last; there are 6 oxygens in glucose and 6 from the water, requiring 6 \(\mathrm{O}_{2}\) for 18 oxygens total. The balanced equation is: \[ 6\mathrm{CO}_{2}(g) + 6\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) + 6\mathrm{O}_{2}(g) \]

Step 3: Balancing Aluminum Sulfate Reaction

For \(\mathrm{Al}(\mathrm{OH})_{3}(s) + \mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(s) + \mathrm{H}_{2} \mathrm{O}(l)\), start with aluminum. Place a coefficient of 2 before \(\mathrm{Al}(\mathrm{OH})_{3}\) to balance 2 aluminum atoms on the right. Next, balance \(\mathrm{SO}_{4}\): you need 3 \(\mathrm{H}_{2} \mathrm{SO}_{4}\) for 3 sulfates in \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\). Finally, balance hydrogen and oxygen with water; 6 water molecules will balance both. The balanced equation is: \[ 2\mathrm{Al}(\mathrm{OH})_{3}(s) + 3\mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}(s) + 6\mathrm{H}_{2} \mathrm{O}(l) \]

Step 4: Balancing Pyrophosphoric Acid Formation

In \(\mathrm{H}_{3} \mathrm{PO}_{4}(aq) \longrightarrow \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}(aq) + \mathrm{H}_{2} \mathrm{O}(l)\), start by balancing phosphorus atoms. Place a 2 in front of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) to get 6 hydrogen and 2 phosphorus atoms, matching the right side. Finally, check oxygen balancing: 8 oxygens in reactant and product match (by considering the water molecule's contribution). The balanced equation is: \[ 2\mathrm{H}_{3} \mathrm{PO}_{4}(aq) \longrightarrow \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \]

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that focuses on the quantitative relationships between the substances involved in chemical reactions. It involves:

• Calculating the relative quantities of reactants and products in a balanced chemical equation.
• Understanding coefficients in a chemical equation, which are used to balance the equation and ensure the law of conservation of mass is respected.

So, when you have a balanced equation, it indicates how many moles of each substance are needed or produced. For instance, in the balanced reaction for the formation of glucose, \[ 6\mathrm{CO}_{2} + 6\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} + 6\mathrm{O}_{2} \],stoichiometry tells us that for every 6 moles of carbon dioxide and 6 moles of water, we produce one mole of glucose and 6 moles of oxygen. This is a reflection of the role that stoichiometry plays in understanding not only the amounts but also the ratios of reactants and products involved in chemical reactions.

Chemical Reactions

Chemical reactions are processes where substances, the reactants, are transformed into different substances, the products. Each equation represents a unique transformation.

• They involve breaking old chemical bonds and forming new ones.
• Energy is either absorbed or released during the reaction.

In the reactions provided, different types of transformations occur:

• In the **Formation of Silicon Hydroxide**, chlorine from silicon tetrachloride reacts with water molecules to produce hydrochloric acid.
• **Glucose Formation** involves photosynthesis-like reactions, showcasing the conversion of carbon dioxide and water into glucose and oxygen.
• The **Reaction with Aluminum Sulfate** involves substitution and combination reactions to form aluminum sulfate from aluminum hydroxide and sulfuric acid.

These examples highlight that each reaction has particular reactants and products, which are specific to the chemical changes. Understanding chemical reactions helps us predict the products and the conditions under which these changes occur.

Conservation of Mass

The principle of conservation of mass states that mass cannot be created or destroyed in a chemical reaction. This principle is crucial when balancing chemical equations:

• Each side of a chemical equation must represent equal mass or quantity of atoms for each element involved.
• This is why balancing equations is essential before proceeding with any further stoichiometric calculations.

For example, in the equation \[ \mathrm{Al} ( \mathrm{OH} )_{3} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} + \mathrm{H}_{2} \mathrm{O} \],each side of the equation has:

• 2 aluminum atoms,
• 3 sulfate ions,
• 6 oxygen and hydrogen atoms provided by water molecules.

No atoms are lost or gained; they are merely rearranged. This verifies that the mass is conserved. By understanding and applying this concept, chemists are able to precisely predict the amounts of substances consumed and produced in chemical reactions, making it a cornerstone of chemistry.