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Write balanced equations based on the information given. (a) solid magnesium \(+\) nitrogen gas \(\longrightarrow\) solid magnesium nitride (b) solid potassium chlorate \(\longrightarrow\) solid potassium chloride + oxygen gas (c) solid sodium hydroxide \(+\) solid ammonium chloride \(\longrightarrow\) solid sodium chloride \(+\) gaseous ammonia \(+\) water vapor (d) solid sodium + liquid water \(\longrightarrow\) aqueous sodium hydroxide \(+\) hydrogen gas

Short Answer

Expert verified
The balanced equations are:\n(a) \(3\text{Mg (s)} + \text{N}_2 (g) \rightarrow \text{Mg}_3 \text{N}_2 (s)\)\n(b) \(2\text{KClO}_3 (s) \rightarrow 2\text{KCl (s)} + 3\text{O}_2 (g)\)\n(c) \(\text{NaOH (s)} + \text{NH}_4\text{Cl (s)} \rightarrow \text{NaCl (s)} + \text{NH}_3 (g) + \text{H}_2\text{O (g)}\)\n(d) \(2\text{Na (s)} + 2\text{H}_2\text{O (l)} \rightarrow 2\text{NaOH (aq)} + \text{H}_2 (g)\)

Step by step solution

01

Equation (a)

Write down reactants and products: \[ \text{Mg (s)} + \text{N}_2 (g) \rightarrow \text{Mg}_3 \text{N}_2 (s) \]. Now balance the equation, resulting in: \[ 3\text{Mg (s)} + \text{N}_2 (g) \rightarrow \text{Mg}_3 \text{N}_2 (s) \]
02

Equation (b)

Write down reactants and products: \[ 2\text{KClO}_3 (s) \rightarrow 2\text{KCl (s)} + 3\text{O}_2 (g) \] The equation is already balanced.
03

Equation (c)

Write down reactants and products: \[ \text{NaOH (s)} + \text{NH}_4\text{Cl (s)} \rightarrow \text{NaCl (s)} + \text{NH}_3 (g) + \text{H}_2\text{O (g)} \] Balance the equation, leading to: \[ \text{NaOH (s)} + \text{NH}_4\text{Cl (s)} \rightarrow \text{NaCl (s)} + \text{NH}_3 (g) + \text{H}_2\text{O (g)} \]
04

Equation (d)

Write down reactants and products: \[ 2\text{Na (s)} + 2\text{H}_2\text{O (l)} \rightarrow 2\text{NaOH (aq)} + \text{H}_2 (g) \] The equation is balanced.

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Key Concepts

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

Stoichiometry
Stoichiometry is a field of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. When dealing with chemical equations, stoichiometry helps to determine the amount of each substance needed or produced. It primarily involves the calculation of moles, masses, volumes, and particles.
Understanding stoichiometry is essential for predicting how substances will react in a specific ratio.
Here's how stoichiometry applies to a balanced chemical equation:
  • It allows us to relate reactants and products using coefficients, which represent a mole ratio.
  • The coefficients are used to calculate the exact quantity of each substance involved in the reaction.
  • This helps in practical scenarios, such as in laboratory experiments, where precise amounts are crucial.
By mastering stoichiometry, you can provide important information about the quantitative aspects of chemical reactions.
Chemical Reactions
A chemical reaction involves the transformation of reactants into products. During this process, original chemical bonds are broken, and new ones are formed, resulting in one or more substances changing into a different one. Every chemical reaction can be represented by a chemical equation that shows both reactants and products.
Chemical reactions are fundamental to understanding chemical processes and have various types based on the rearrangement of atoms. Some are:
  • Synthesis reactions: Two or more simple substances combine to form a complex product, as seen in the reaction of magnesium with nitrogen to form magnesium nitride.
  • Decomposition reactions: A complex molecule breaks down into simpler ones, like potassium chlorate producing potassium chloride and oxygen.
Recognizing different types of reactions is crucial for predicting the outcomes in practical and industrial applications.
Chemical Equations
Chemical equations are symbolic representations of chemical reactions. They consist of chemical formulas that represent the reactants and products involved. A chemical equation not only indicates the substances but also the relative amounts needed for the reaction to proceed.
An essential aspect of chemical equations is their ability to clearly show the state of each substance (solid, liquid, gas, or aqueous) and provide a visual map of a chemical change.
Key components include:
  • Reactants: Starting materials at the left.
  • Products: Final substances at the right.
  • Coefficients: Numbers indicating the proportion of each substance.
When writing chemical equations, ensuring they are balanced is vital for maintaining the law of conservation of mass.
Reaction Balancing Methods
Balancing chemical equations is crucial for maintaining the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. The goal is to have the same number of each type of atom on both sides of the equation.
Here are some common methods for balancing equations:
  • Inspection method: Systematically adjusting coefficients to balance each element.
  • Algebraic method: Using mathematical equations to solve for unknown coefficients.

When balancing equations as illustrated in the example problems:
  • Start by writing an unbalanced equation with the given reactants and products.
  • Begin with the most complex molecule and systematically address each element.
  • Adjust coefficients iteratively until balance is achieved, ensuring that the same quantity of each atom exists on both sides of the equation.
This practice not only ensures compliance with physical laws but also helps in accurate stoichiometric calculations.

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Most popular questions from this chapter

How many milligrams \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) must be present in \(50.0 \mathrm{L}\) of a solution containing \(2.35 \mathrm{ppm} \mathrm{Ca} ?\) [Hint: See also Exercise 103 .]

How many grams of sodium must react with \(155 \mathrm{mL}\) \(\mathrm{H}_{2} \mathrm{O}\) to produce a solution that is \(0.175 \mathrm{M} \mathrm{NaOH} ?\) (Assume a final solution volume of \(155 \mathrm{mL}\) ) $$ 2 \mathrm{Na}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) $$

Baking soda, \(\mathrm{NaHCO}_{3}\), is made from soda ash, a common name for sodium carbonate. The soda ash is obtained in two ways. It can be manufactured in a process in which carbon dioxide, ammonia, sodium chloride, and water are the starting materials. Alternatively, it is mined as a mineral called trona (left photo). Whether the soda ash is mined or manufactured, it is dissolved in water and carbon dioxide is bubbled through the solution. Sodium bicarbonate precipitates from the solution. As a chemical analyst you are presented with two samples of sodium bicarbonate-one from the manufacturing process and the other derived from trona. You are asked to determine which is purer and are told that the impurity is sodium carbonate. You decide to treat the samples with just sufficient hydrochloric acid to convert all the sodium carbonate and bicarbonate to sodium chloride, carbon dioxide, and water. You then precipitate silver chloride in the reaction of sodium chloride with silver nitrate. A \(6.93 \mathrm{g}\) sample of baking soda derived from trona gave \(11.89 \mathrm{g}\) of silver chloride. A \(6.78 \mathrm{g}\) sample from manufactured sodium carbonate gave \(11.77 \mathrm{g}\) of silver chloride. Which sample is purer, that is, which has the greater mass percent \(\mathrm{NaHCO}_{3} ?\)

Briefly describe (a) balancing a chemical equation; (b) preparing a solution by dilution; (c) determining the limiting reactant in a reaction.

Suppose that reactions (a) and (b) each have a \(92 \%\) yield. Starting with \(112 \mathrm{g} \mathrm{CH}_{4}\) in reaction \((\mathrm{a})\) and an excess of \(\mathrm{Cl}_{2}(\mathrm{g}),\) how many grams of \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are formed in reaction (b)? (a) \(\mathrm{CH}_{4}+\mathrm{Cl}_{2} \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}+\mathrm{HCl}\) (b) \(\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{CH}_{2} \mathrm{Cl}_{2}+\mathrm{HCl}\)

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