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Complete and balance the following equations: (a) \(\mathrm{CO}_{2}(g)+\mathrm{OH}^{-}(a q) \longrightarrow\) (b) \(\mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow\) (c) \(\mathrm{CaO}(s)+\mathrm{C}(s) \stackrel{\Delta}{\longrightarrow}\) (d) \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \stackrel{\Delta}{\longrightarrow}\) (e) \(\mathrm{CuO}(s)+\mathrm{CO}(g)\)

Short Answer

Expert verified
(a) CO鈧(g) + 2OH鈦(aq) 鉄 H鈧侽(l) + CO鈧兟测伝(aq) (b) NaHCO鈧(s) + H鈦(aq) 鉄 Na鈦(aq) + H鈧侽(l) + CO鈧(g) (c) CaO(s) + C(s) 鉄 CaC鈧(s) + CO(g) (d) C(s) + H鈧侽(g) 鉄 CO(g) + H鈧(g) (e) CuO(s) + CO(g) 鉄 Cu(s) + CO鈧(g)

Step by step solution

01

(a) Strategy to balance the equation

To balance the equation, identify the ions and elements on both sides of the equation. Write the balanced form of the equation, ensuring that there are equal numbers of each element on both the reactant and product sides. In this case: CO鈧(g) + OH鈦(aq) 鉄
02

(a) Balancing the equation

To balance the equation, we need equal numbers of each element and ion on both sides. After analyzing, we can write the balanced equation as: CO鈧(g) + 2OH鈦(aq) 鉄 H鈧侽(l) + CO鈧兟测伝(aq)
03

(b) Strategy to balance the equation

Following the same strategy, we'll identify the ions and elements on both sides of the equation. Then, write the balanced form of the equation: NaHCO鈧(s) + H鈦(aq) 鉄
04

(b) Balancing the equation

To balance the given equation, we get: NaHCO鈧(s) + H鈦(aq) 鉄 Na鈦(aq) + H鈧侽(l) + CO鈧(g)
05

(c) Strategy to balance the equation

Repeating the strategy for equation (c): CaO(s) + C(s) 鉄
06

(c) Balancing the equation

Balance the equation to obtain: CaO(s) + C(s) 鉄 CaC鈧(s) + CO(g)
07

(d) Strategy to balance the equation

Once again, apply the strategy to balance the equation: C(s) + H鈧侽(g) 鉄
08

(d) Balancing the equation

Balancing the given equation, we find: C(s) + H鈧侽(g) 鉄 CO(g) + H鈧(g)
09

(e) Strategy to balance the equation

Applying the same strategy for the last equation: CuO(s) + CO(g) 鉄
10

(e) Balancing the equation

Balance the equation to get: CuO(s) + CO(g) 鉄 Cu(s) + CO鈧(g)

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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 reactants transform into products, often by breaking and forming bonds. Each reaction is subject to the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. This principle is foundational to understanding how to balance chemical equations.

For example, in the reaction between carbon dioxide (CO鈧) and hydroxide ions (OH鈦), the products are water (H鈧侽) and carbonate ions (CO鈧兟测伝). Reactants are on the left, products on the right, and the arrow signifies the direction of transformation. The purpose of balancing is to ensure that the number of atoms for each element is the same on both sides, aligning with the law of conservation of mass.
Stoichiometry
Stoichiometry comes from the Greek words for 'element' and 'measure' and is a quantitative relationship between the reactants and products in a chemical reaction. It is built upon the idea that relationships in chemistry are governed by ratios. These ratios are determined by stoichiometric coefficients, the numbers placed in front of compounds in a chemical equation.

The balanced chemical equation for the reaction of sodium bicarbonate (NaHCO鈧) with hydrogen ions (H鈦) demonstrates stoichiometry. The equation NaHCO鈧(s) + H鈦(aq) 鉄 Na鈦(aq) + H鈧侽(l) + CO鈧(g) shows a 1:1 ratio between sodium bicarbonate and hydrogen ions, ensuring mass conservation and providing a basis for calculations involving the quantities of reactants and products.
Reaction Balancing Strategies
To balance a chemical equation, certain strategies can guide you through the sometimes complex process. Common steps include first balancing elements that appear in only one reactant and one product, balancing more complex molecules before simpler ones, and leaving hydrogen and oxygen to be balanced last due to their frequent occurrence.

In the given example of balancing calcium oxide (CaO) reacting with carbon (C) to produce calcium carbide (CaC鈧) and carbon monoxide (CO), the strategy was identifying and balancing compounds based on their occurrence in the equation. Once CaC鈧 and CO were determined as products, it made clear what reactants must be present in equal numbers to obey the conservation of mass. It's helpful to approach balancing with a systematic method, adjusting coefficients incrementally until equilibrium is achieved.

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Most popular questions from this chapter

(a) The \(\mathrm{P}_{4}, \mathrm{P}_{4} \mathrm{O}_{6}\) and \(\mathrm{P}_{4} \mathrm{O}_{10}\) molecules have a common structural feature of four \(\mathrm{P}\) atoms arranged in a tetrahedron (Figures \(22.27\) and 22.28). Does this mean that the bonding between the \(\mathrm{P}\) atoms is the same in all these cases? Explain. (b) Sodium trimetaphosphate \(\left(\mathrm{Na}_{3} \mathrm{P}_{3} \mathrm{O}_{9}\right)\) and sodium tetrametaphosphate \(\left(\mathrm{Na}_{4} \mathrm{P}_{4} \mathrm{O}_{12}\right)\) are used as water-softening agents. They contain cyclic \(\mathrm{P}_{3} \mathrm{O}_{9}{ }^{3-}\) and \(\mathrm{P}_{4} \mathrm{O}_{12}{ }^{4-}\) ions, respectively. Propose reasonable structures for these ions.

SOLUTION Analyze We are asked to determine the nucleus that results when radium-226 loses an alpha particle. Plan We can best do this by writing a balanced nuclear reaction for the process. Solve The periodic table shows that radium has an atomic number of 88 . The complete chemical symbol for radium- 226 is therefore \({ }_{85}^{226} \mathrm{Ra}\). An alpha particle is a helium-4 nucleus, and so its symbol is \({ }_{2}^{4} \mathrm{He}\). The alpha particle is a product of the nuclear reaction, and so the equation is of the form $$ { }_{8}^{226} \mathrm{Ra} \longrightarrow{ }_{2}^{A} \mathrm{X}+{ }_{2}^{4} \mathrm{He} $$ where \(A\) is the mass number of the product nucleus and \(Z\) is its atomic number. Mass numbers and atomic numbers must balance, so $$ 226=A+4 $$ and $$ 88=Z+2 $$ Hence, $$ A=222 \text { and } Z=86 $$ Again, from the periodic table, the element with \(Z=86\) is radon (Rn). The product, therefore, is \({ }_{86}^{222} \mathrm{Rn}\), and the nuclear equation is $$ { }_{88}^{226} \mathrm{Ra} \longrightarrow{ }_{86}^{222} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He} $$

Give a reason why hydrogen might be placed along with the group lA elements of the periodic table.

Account for the following observations: (a) \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is a diprotic acid. (b) Nitric acid is a strong acid, whereas phosphoric acid is weak. (c) Phosphate rock is ineffective as a phosphate fertilizer. (d) Phosphorus does not exist at room temperature as diatomic molecules, but nitrogen does. (e) Solutions of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) are quite basic.

Hydrazine has been employed as a reducing agent for metals. Using standard reduction potentials, predict whether the following metals can be reduced to the metallic state by hydrazine under standard conditions in acidic solution: (a) \(\mathrm{Fe}^{2+}\), (b) \(\mathrm{Sn}^{2+}\), (c) \(\mathrm{Cu}^{2+}\), (d) \(\mathrm{Ag}^{+}\), (e) \(\mathrm{Cr}^{3+}\), (f) \(\mathrm{Co}^{3+}\).

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