Problem 66 - Chemistry Matter and Change

Question

Balance the following reactions. $$\begin{array}{l}{\text { a. }\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{s}) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})} \\ {\text { b. } \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(1) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g})}\end{array}$$

Step-by-Step Solution

Answer

(a) \((NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3 + N_2 + 4H_2O\); (b) \(6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2\).

Step 1: Identify the Components of Reaction (a)

In the reaction \((NH_4)_2Cr_2O_7(s) \rightarrow Cr_2O_3(s) + N_2(g) + H_2O(g)\), note the elements present, which are nitrogen \((N)\), hydrogen \((H)\), chromium \((Cr)\), and oxygen \((O)\). We need the same number of each element on both sides of the reaction.

Step 2: Balance Chromium in Reaction (a)

In the reactant \((NH_4)_2Cr_2O_7\), there are 2 chromium atoms present. Ensure there are exactly 2 chromium atoms in the product too. The product \(Cr_2O_3\) already contains 2 chromium atoms, so chromium is balanced.

Step 3: Balance Nitrogen in Reaction (a)

There are 2 nitrogen atoms in \((NH_4)_2Cr_2O_7\) as there are two ammonium \((NH_4)^+\) ions. This means that the \(N_2\) gas in the products is balanced as it also contains 2 nitrogen atoms.

Step 4: Balance Hydrogen in Reaction (a)

There are 8 hydrogen atoms in \((NH_4)_2Cr_2O_7\) (from the two ammonium ions). As each water molecule \(H_2O\) contains 2 hydrogen atoms, we need 4 water molecules to balance the hydrogen: \(4H_2O\).

Step 5: Balance Oxygen in Reaction (a)

There are 7 oxygen atoms in \((NH_4)_2Cr_2O_7\). In the products: \(Cr_2O_3\) has 3 oxygen atoms and \(4H_2O\) contribute 4 oxygen atoms, totaling 7 oxygen atoms. Therefore, oxygen is balanced.

Step 6: Balanced Equation for Reaction (a)

The balanced reaction for (a) is: \[(NH_4)_2Cr_2O_7(s) \rightarrow Cr_2O_3(s) + N_2(g) + 4H_2O(g)\].

Step 7: Identify the Components of Reaction (b)

For reaction \(CO_2(g) + H_2O(l) \rightarrow C_6H_{12}O_6(s) + O_2(g)\), identify carbon \((C)\), hydrogen \((H)\), and oxygen \((O)\) as the present elements.

Step 8: Balance Carbon in Reaction (b)

The molecule \(C_6H_{12}O_6\) has 6 carbon atoms. To balance carbon, 6 \(CO_2\) molecules are needed on the reactant side: \[6CO_2\].

Step 9: Balance Hydrogen in Reaction (b)

The molecule \(C_6H_{12}O_6\) contains 12 hydrogen atoms, so we need 6 \(H_2O\) molecules to balance the hydrogen on the reactant side: \[6H_2O\].

Step 10: Balance Oxygen in Reaction (b)

On the reactants side, there are 12 oxygen atoms in \(6CO_2\) and 6 in \(6H_2O\), totaling 18. In the products, \(C_6H_{12}O_6\) has 6 oxygen atoms. Thus, 6 \(O_2\) molecules provide the other 12 oxygen atoms needed to balance: \[6O_2\].

Step 11: Balanced Equation for Reaction (b)

The balanced reaction for (b) is: \[6CO_2(g) + 6H_2O(l) \rightarrow C_6H_{12}O_6(s) + 6O_2(g)\].

Key Concepts

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It's a key aspect of chemistry that helps us measure the exact amounts of substances involved in reactions. This concept is crucial because chemical reactions must obey the Law of Conservation of Mass, which we'll get into later.

By using stoichiometry, we can determine how much of each reactant is needed to form a desired amount of product. For example, in the combustion of hydrogen with oxygen, stoichiometry tells us that two moles of hydrogen gas react with one mole of oxygen gas to form two moles of water. This is reflected in the balanced chemical equation:

• \[2H_2(g) + O_2(g) \rightarrow 2H_2O(l)\]

In the exercises provided, we applied stoichiometry by counting the number of atoms of each element on both sides of the reaction. By doing so, we ensured that the reactions were balanced, meaning the number of each type of atom was the same on both sides.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances, called products. These transformations involve the making and breaking of chemical bonds, leading to new substances with different properties than the original reactants.

In our exercises, we examined two types of reactions:

• Decomposition Reaction: Reaction (a) involved the breakdown of ammonium dichromate \((NH_4)_2Cr_2O_7\) into chromium oxide \(Cr_2O_3\), nitrogen gas \(N_2\), and water \(H_2O\).
• Photosynthesis-like Reaction: Reaction (b) involved the combination of carbon dioxide \(CO_2\) and water \(H_2O\) to form glucose \(C_6H_{12}O_6\) and oxygen gas \(O_2\).

Understanding chemical reactions is key to predicting how different substances will interact and transform in various conditions.

Chemical Equation Balancing

Chemical equation balancing is a fundamental process in chemistry used to ensure the same number of each type of atom appears on both the reactants and products sides of a reaction. This reflects the principle that mass cannot be created or destroyed in a chemical reaction.

The steps to balance chemical equations begin with:

• Identifying the elements in the reaction.
• Counting the number of atoms for each element on both sides of the equation.
• Adjusting coefficients – the numbers in front of molecules – to make sure each side has the same number of atoms for every element.

For our reactions, balancing involved making the number of carbon, hydrogen, chromium, nitrogen, and oxygen atoms equal on both sides of each equation. This ensures that every single atom in the reactants is accounted for in the products.

Law of Conservation of Mass

The Law of Conservation of Mass states that mass in a closed system cannot be created or destroyed. It implies that the mass of reactants must equal the mass of products in a chemical reaction.

This law is why balancing chemical equations is necessary. For instance, when balancing reaction (a), we ensured the reactant side and the product side had equal numbers of each type of atom. This balancing act keeps the mass constant throughout the reaction.
A classic example demonstrating this law is the reaction of hydrogen with oxygen to form water:

• \[2H_2(g) + O_2(g) \rightarrow 2H_2O(l)\]

From start to finish, every hydrogen and oxygen atom from the reactants are found in the water produced, maintaining mass consistent throughout the process.