Problem 46 - Chemistry and Chemical Reactivity

Question

Balance the following equations: (a) Reaction to produce "superphosphate" fertilizer \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}(\mathrm{aq})+\mathrm{CaSO}_{4}(\mathrm{s})\) (b) Reaction to produce diborane, \(B_{2} H_{6}\) \(\mathrm{NaBH}_{4}(\mathrm{s})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})+\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) (c) Reaction to produce tungsten metal from tungsten (VI) oxide \(\mathrm{WO}_{3}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{W}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (d) Decomposition of ammonium dichromate \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{s}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})\)

Step-by-Step Solution

Answer

(a) \(\mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} + 3\mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow 2\mathrm{Ca}(\mathrm{H}_{2}\mathrm{PO}_{4})_{2} + \mathrm{CaSO}_{4}\) (b) \(2\mathrm{NaBH}_{4} + \mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{B}_{2}\mathrm{H}_{6} + 2\mathrm{H}_{2} + \mathrm{Na}_{2}\mathrm{SO}_{4}\) (c) \(\mathrm{WO}_{3} + 3\mathrm{H}_{2} \rightarrow \mathrm{W} + 3\mathrm{H}_{2}\mathrm{O}\) (d) \((\mathrm{NH}_{4})_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} \rightarrow \mathrm{N}_{2} + 4\mathrm{H}_{2}\mathrm{O} + \mathrm{Cr}_{2}\mathrm{O}_{3}\)

Step 1: Analyzing Equation (a) for Elements

In equation (a), identify the elements present: Calcium (Ca), Phosphorus (P), and Sulfur (S), along with Hydrogen (H) and Oxygen (O). Each side of the equation needs an equal number of each atom.

Step 2: Balancing Equation (a)

Start by balancing Phosphorus. There are 2 P atoms in \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\), so two \(\mathrm{Ca}\left(\mathrm{H}_{2}\mathrm{PO}_{4}\right)_{2}\) are needed. Next, Calcium: 3 Ca atoms in the reactant, so ensure 3 in products: 1 \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) + 2 \(\mathrm{CaSO}_{4}\). Balance Sulfur with 3 \(\mathrm{H}_{2}\mathrm{SO}_{4}\) to match 3 \(\mathrm{CaSO}_{4}\). Balancing gives: \[\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+3\mathrm{H}_{2}\mathrm{SO}_{4} \longrightarrow 2 \mathrm{Ca}\left(\mathrm{H}_{2}\mathrm{PO}_{4}\right)_{2}+\mathrm{CaSO}_{4}\]

Step 3: Analyzing Equation (b) for Elements

Identify elements in equation (b): Sodium (Na), Boron (B), Hydrogen (H), and Sulfur (S), along with Oxygen (O). The number of each type of atom must be equal on both sides.

Step 4: Balancing Equation (b)

Balance Boron first using 2 \(\mathrm{NaBH}_{4}\), ensuring 2 B atoms. Adjust Sodium with \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) as needed. Balance Hydrogen by having 5 \(\mathrm{H}_{2}\) molecules, ensuring balance of Hydrogen with reactants and products. Then check for Oxygen. Balanced equation: \[2 \mathrm{NaBH}_{4}+\mathrm{H}_{2}\mathrm{SO}_{4} \longrightarrow \mathrm{B}_{2}\mathrm{H}_{6}+2 \mathrm{H}_{2}+\mathrm{Na}_{2}\mathrm{SO}_{4}\]

Step 5: Analyzing Equation (c) for Elements

For equation (c), consider elements Tungsten (W), Hydrogen (H), and Oxygen (O). Each element should have the same number of atoms on both sides.

Step 6: Balancing Equation (c)

With 1 W atom on both sides, balance Hydrogen by ensuring 3 \(\mathrm{H}_{2}\) molecules provide 6 H atoms total. Match Oxygen atoms with 3 \(\mathrm{H}_{2}\mathrm{O}\) yielding 3 \(\mathrm{O}\) atoms. The balanced equation is: \[\mathrm{WO}_{3}+3\mathrm{H}_{2} \longrightarrow \mathrm{W}+3\mathrm{H}_{2}\mathrm{O}\]

Step 7: Analyzing Equation (d) for Elements

Identify elements in equation (d): Nitrogen (N), Hydrogen (H), Chromium (Cr), and Oxygen (O). Verify atom count equality on both sides.

Step 8: Balancing Equation (d)

Start by balancing Chromium directly with \(\mathrm{Cr}_{2}\mathrm{O}_{3}\), followed by balancing Nitrogen using 1 \(\mathrm{N}_{2}\). Adjust Hydrogen by ensuring 4 \(\mathrm{H}_{2}\mathrm{O}\) provides the correct count, and then ensure Oxygen is balanced. The balanced equation: \[\left(\mathrm{NH}_{4}\right)_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} \longrightarrow \mathrm{N}_{2}+4\mathrm{H}_{2}\mathrm{O}+\mathrm{Cr}_{2}\mathrm{O}_{3}\]

Key Concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances, called products. This transformation involves rearranging the atoms within the molecules to form new bonds. During balancing these chemical equations, we aim to have an equal number of each type of atom on both sides. This is because, according to the law of conservation of mass, matter cannot be created or destroyed in a closed system. We explored this through reactions like the production of superphosphate fertilizer, which requires us to correctly balance elements such as calcium (Ca), phosphorus (P), sulfur (S), hydrogen (H), and oxygen (O). In another equation, the manufacturing of tungsten metal involves converting tungsten (VI) oxide (WO extsubscript{3}) into tungsten (W) utilizing hydrogen (H extsubscript{2}). Here again, we ensure the atoms of tungsten, hydrogen, and oxygen are equally represented on both sides. Each balanced chemical equation represents a snapshot of the detailed molecular changes during a reaction.

Stoichiometry

Stoichiometry is essential for understanding chemical reactions, as it involves calculating the exact relationships between reactants and products. In any chemical reaction, stoichiometry allows us to predict how much of each substance is consumed or produced. Take, for instance, the creation of diborane (B extsubscript{2}H extsubscript{6}) from sodium borohydride (NaBH extsubscript{4}) and sulfuric acid (H extsubscript{2}SO extsubscript{4}). The stoichiometric coefficients indicate the ratios needed for complete reactions. In this balanced chemical equation: - 2 NaBH extsubscript{4} react with 1 H extsubscript{2}SO extsubscript{4}, producing 1 B extsubscript{2}H extsubscript{6}, 2 H extsubscript{2}, and 1 Na extsubscript{2}SO extsubscript{4}. Understanding these coefficients helps predict outcomes such as the quantity of gas formed or reactants required. This is crucial for both lab-based experiments and industrial applications where precise quantities matter for efficiency and economy.

Chemical Formulas

Chemical formulas are representations of molecules using element symbols and numerical subscripts. They provide insights into the composition and structure of compounds. For example, in the decomposition of ammonium dichromate (NH extsubscript{4}) extsubscript{2}Cr extsubscript{2}O extsubscript{7} breaks down into nitrogen gas (N extsubscript{2}), water (H extsubscript{2}O), and chromium (III) oxide (Cr extsubscript{2}O extsubscript{3}). The chemical formula denotes the exact number of atoms in a molecule. In ammonium dichromate: - 2 ammonium (NH extsubscript{4}) units indicate 8 hydrogen atoms (H) and 2 nitrogen atoms (N). - Each Cr extsubscript{2}O extsubscript{7} unit consists of 2 chromium atoms (Cr) and 7 oxygen atoms (O). Understanding chemical formulas is crucial, as they inform how we balance equations by depicting how molecules react and what they produce. This knowledge is foundational for accurately interpreting and predicting chemical behaviors.