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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 \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{s}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})\)

Short Answer

Expert verified
(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 by step solution

01

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.
02

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: \[\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+3\mathrm{H}_{2}\mathrm{SO}_{4} \longrightarrow 2 \mathrm{Ca}\left(\mathrm{H}_{2}\mathrm{PO}_{4}\right)_{2}+\mathrm{CaSO}_{4}\]
03

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.
04

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: \[2 \mathrm{NaBH}_{4}+\mathrm{H}_{2}\mathrm{SO}_{4} \longrightarrow \mathrm{B}_{2}\mathrm{H}_{6}+2 \mathrm{H}_{2}+\mathrm{Na}_{2}\mathrm{SO}_{4}\]
05

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.
06

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: \[\mathrm{WO}_{3}+3\mathrm{H}_{2} \longrightarrow \mathrm{W}+3\mathrm{H}_{2}\mathrm{O}\]
07

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.
08

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: \[\left(\mathrm{NH}_{4}\right)_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} \longrightarrow \mathrm{N}_{2}+4\mathrm{H}_{2}\mathrm{O}+\mathrm{Cr}_{2}\mathrm{O}_{3}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Your body deals with excess nitrogen by excreting it in the form of urea, \(\mathrm{NH}_{2} \mathrm{CONH}_{2}\). The reaction producing it is the combination of arginine \(\left(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{N}_{4} \mathrm{O}_{2}\right)\) with water to give urea and ornithine \(\left(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{2}\right)\) \(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{N}_{4} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}+\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{2}\) Arginine Urea Ornithine If you excrete 95 mg of urea, what mass of arginine must have been used? What mass of ornithine must have been produced?

To find the formula of a compound composed of iron and carbon monoxide, \(\mathrm{Fe}_{x}(\mathrm{CO})_{y}\), the compound is burned in pure oxygen to give \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(\mathrm{CO}_{2} .\) If you burn \(1.959 \mathrm{g}\) of \(\mathrm{Fe}_{x}(\mathrm{CO})_{y}\) and obtain \(0.799 \mathrm{g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(2.200 \mathrm{g}\) of \(\mathrm{CO}_{2}\) what is the empirical formula of \(\mathrm{Fe}_{x}(\mathrm{CO})_{y^{2}}\)

A reaction studied by Wächtershäuser and Huber (see "Black Smokers and the Origins of Life") is $$ 2 \mathrm{CH}_{3} \mathrm{SH}+\mathrm{CO} \longrightarrow \mathrm{CH}_{3} \mathrm{COSCH}_{3}+\mathrm{H}_{2} \mathrm{S} $$ If you begin with \(10.0 \mathrm{g}\) of \(\mathrm{CH}_{3} \mathrm{SH}\), and excess \(\mathrm{CO}\), (a) What is the theoretical yield of \(\mathrm{CH}_{3} \mathrm{COSCH}_{3} ?\) (b) If 8.65 g of \(\mathrm{CH}_{3} \mathrm{COSCH}_{3}\) is isolated, what is its percent yield?

Ammonia gas can be prepared by the following reaction: \(\mathrm{CaO}(\mathrm{s})+2 \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CaCl}_{2}(\mathrm{s})\) If \(112 \mathrm{g}\) of \(\mathrm{CaO}\) and \(224 \mathrm{g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) are mixed, the theoretical yield of \(\mathrm{NH}_{3}\) is \(68.0 \mathrm{g}\) (Study Question 20 ). If only \(16.3 \mathrm{g}\) of \(\mathrm{NH}_{3}\) is actually obtained, what is its percent yield?

Nickel forms a compound with carbon monoxide, \(\mathrm{Ni}_{x}(\mathrm{CO})_{r}\). To determine its formula, you carefully heat a 0.0973 -g sample in air to convert the nickel to \(0.0426 \mathrm{g}\) of NiO and the CO to \(0.100 \mathrm{g}\) of \(\mathrm{CO}_{2} .\) What is the empirical formula of \(\mathrm{Ni}_{x}(\mathrm{CO})_{y} ?\)

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