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Write an unbalanced equation to represent each of the following reactions: (a) potassium hydroxide and phosphoric acid react to form potassium phosphate and water; \((\mathrm{b})\) zinc and silver chloride react to form zinc chloride and silver; (c) sodium hydrogen carbonate reacts to form sodium carbonate, water, and carbon dioxide; (d) ammonium nitrite reacts to form nitrogen and water; and (e) carbon dioxide and potassium hydroxide react to form potassium carbonate and water. (f) Balance equations (a)-(e).

Short Answer

Expert verified
(a) 3KOH + H₃PO₄ → K₃PO₄ + 3H₂O; (b) Zn + 2AgCl → ZnCl₂ + 2Ag; (c) 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂; (d) NH₄NO₂ → N₂ + 2H₂O; (e) CO₂ + 2KOH → K₂CO₃ + H₂O.

Step by step solution

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01

Write Unbalanced Equation for Reaction (a)

Reaction between potassium hydroxide (KOH) and phosphoric acid (H₃PO₄) to form potassium phosphate (K₃PO₄) and water (H₂O). The unbalanced equation is:\[ \text{KOH} + \text{H}_3\text{PO}_4 \rightarrow \text{K}_3\text{PO}_4 + \text{H}_2\text{O} \]
02

Write Unbalanced Equation for Reaction (b)

Zinc (Zn) reacts with silver chloride (AgCl) to form zinc chloride (ZnCl₂) and silver (Ag). The unbalanced equation is:\[ \text{Zn} + \text{AgCl} \rightarrow \text{ZnCl}_2 + \text{Ag} \]
03

Write Unbalanced Equation for Reaction (c)

Sodium hydrogen carbonate (NaHCO₃) decomposes to form sodium carbonate (Na₂CO₃), water (H₂O), and carbon dioxide (CO₂). The unbalanced equation is:\[ \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \]
04

Write Unbalanced Equation for Reaction (d)

Ammonium nitrite (NH₄NO₂) decomposes to form nitrogen (N₂) and water (H₂O). The unbalanced equation is:\[ \text{NH}_4\text{NO}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O} \]
05

Write Unbalanced Equation for Reaction (e)

Carbon dioxide (CO₂) reacts with potassium hydroxide (KOH) to form potassium carbonate (K₂CO₃) and water (H₂O). The unbalanced equation is:\[ \text{CO}_2 + \text{KOH} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \]
06

Balance Equation (a)

Balance equation for reaction (a):\[ 3\text{KOH} + \text{H}_3\text{PO}_4 \rightarrow \text{K}_3\text{PO}_4 + 3\text{H}_2\text{O} \]
07

Balance Equation (b)

Balance equation for reaction (b):\[ \text{Zn} + 2\text{AgCl} \rightarrow \text{ZnCl}_2 + 2\text{Ag} \]
08

Balance Equation (c)

Balance equation for reaction (c):\[ 2\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \]
09

Balance Equation (d)

Balance equation for reaction (d):\[ \text{NH}_4\text{NO}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O} \]
10

Balance Equation (e)

Balance equation for reaction (e):\[ \text{CO}_2 + 2\text{KOH} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \]

Key Concepts

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

Balancing Chemical Equations
Balancing chemical equations is a crucial step in solving any chemical reaction problem. It ensures that the same number of each type of atom is present on both sides of an equation. This law is based on the conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

When balancing equations, follow these steps:
  • Write down the unbalanced equation with all reactants and products.
  • Count the number of each type of atom on both sides of the equation.
  • Add coefficients (whole numbers) in front of chemical formulas to balance the atom count. Remember, coefficients multiply all the atoms in a formula.
  • Check to ensure all atoms balance across the equation.
For example, to balance the equation \( \text{KOH} + \text{H}_3\text{PO}_4 \rightarrow \text{K}_3\text{PO}_4 + \text{H}_2\text{O} \), it becomes \( 3\text{KOH} + \text{H}_3\text{PO}_4 \rightarrow \text{K}_3\text{PO}_4 + 3\text{H}_2\text{O} \) after balancing.
Reaction Types
Different reaction types are categorized based on the transformation occurring during the reaction. Understanding these types is important because it helps predict the products of a reaction and the conditions needed.
  • Synthesis Reaction: Two or more reactants combine to form a single product.
  • Decomposition Reaction: A single compound breaks down into two or more products.
  • Single Replacement Reaction: An element replaces another in a compound.
  • Double Replacement Reaction: Exchange of ions between two compounds forming new compounds.
Identifying the reaction type is a key part of writing and balancing chemical equations.
Decomposition Reactions
Decomposition reactions occur when a single compound breaks down into two or more simpler substances. These reactions are often induced by energy input such as heat, light, or electricity.

For example, in the decomposition of sodium hydrogen carbonate, which is written as \( \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \), each molecule of sodium hydrogen carbonate decomposes to yield sodium carbonate, water, and carbon dioxide.

These reactions are easily recognizable because they start with one reactant and end with multiple products.
Double Displacement Reactions
Double displacement reactions involve the exchange of ions between two reacting compounds to form two new compounds. This type of reaction often results in the formation of a precipitate, a gas, or water.

For instance, in the reaction between potassium hydroxide and phosphoric acid, represented by the balanced equation \( 3\text{KOH} + \text{H}_3\text{PO}_4 \rightarrow \text{K}_3\text{PO}_4 + 3\text{H}_2\text{O} \), potassium ions trade places with hydrogen ions to form potassium phosphate and water.

Double displacement reactions are common in solutions where ions are free to move and exchange.

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Most popular questions from this chapter

Which of the following has the greater mass: \(0.72 \mathrm{~g}\) of \(\mathrm{O}_{2}\) or \(0.0011 \mathrm{~mol}\) of chlorophyll \(\left(\mathrm{C}_{55} \mathrm{H}_{72} \mathrm{MgN}_{4} \mathrm{O}_{5}\right) ?\)

Hemoglobin \(\left(\mathrm{C}_{2952} \mathrm{H}_{4664} \mathrm{~N}_{812} \mathrm{O}_{832} \mathrm{~S}_{8} \mathrm{Fe}_{4}\right)\) is the oxygen carrier in blood. (a) Calculate its molar mass. (b) An average adult has about \(5.0 \mathrm{~L}\) of blood. Every milliliter of blood has approximately \(5.0 \times 10^{9}\) erythrocytes, or red blood cells, and every red blood cell has about \(2.8 \times 10^{8}\) hemoglobin molecules. Calculate the mass of hemoglobin molecules in grams in an average adult.

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